Probability

Fitting Data

Each permutation of the n M's and k-1 P's corresponds to a particular distribution of n M's among k boxes. Where there are two or more P's right next to each other, there is an empty box between the P's.

The number of permutations of n M's and k-1 P's is equal to the number of permutations of n+k-1 objects where n are of one type and k-1 are of another type. This is the number of ways n marbles can be distributed among k boxes,

N(n, k) = (n+k-1)!/[n!(k-1)!]

N(5, 3) = (5+3-1)!/[5!(3-1)!] = 7!/(5!2!) = 21

There are 21 ways to distribute 5 indistinguishable marbles among 3 boxes. These are illustrated in the table below, which shows the numbers of marbles in each box, for all 21 possible arrangements, and the corresponding M and P representations, as discussed above.

In filling up this table, we first consider the case where all 5 marbles are in a single box. Then we consider the cases where the 5 marbles are spread among 2 boxes. Finally, we consider the cases where the 5 marbles are spread among all 3 boxes.

Case | Box 1 | Box 2 | Box 3 | M and P Representation |
---|---|---|---|---|

1 |
5 |
0 |
0 |
MMMMMPP |

2 |
0 |
5 |
0 |
PMMMMMP |

3 |
0 |
0 |
5 |
PPMMMMM |

4 |
4 |
1 |
0 |
MMMMPMP |

5 |
4 |
0 |
1 |
MMMMPPM |

6 |
1 |
4 |
0 |
MPMMMMP |

7 |
0 |
4 |
1 |
PMMMMPM |

8 |
1 |
0 |
4 |
MPPMMMM |

9 |
0 |
1 |
4 |
PMPMMMM |

10 |
3 |
2 |
0 |
MMMPMMP |

11 |
3 |
0 |
2 |
MMMPPMM |

12 |
2 |
3 |
0 |
MMPMMMP |

13 |
0 |
3 |
2 |
PMMMPMM |

14 |
2 |
0 |
3 |
MMPPMMM |

15 |
0 |
2 |
3 |
PMMPMMM |

16 |
3 |
1 |
1 |
MMMPMPM |

17 |
1 |
3 |
1 |
MPMMMPM |

18 |
1 |
1 |
3 |
MPMPMMM |

19 |
2 |
2 |
1 |
MMPMMPM |

20 |
2 |
1 |
2 |
MMPMPMM |

21 |
1 |
2 |
2 |
MPMMPMM |

In how many ways can 7 books be arranged on a shelf if (a) any arrangement is possible, (b) 3 particular books must always stand together, (c) two particular books must occupy the ends?

[from Spiegel, Murray R. 1975,

SOLUTION

(a) If any arrangement is possible, there are 7! = 5040 ways to arrange 7 books on a shelf.

(b) If 3 particular books must always stand together, the 3 books can occupy the following positions on the shelf:

XXX----

-XXX---

--XXX--

---XXX-

----XXX

For each of these five cases, there are 3! permutations of the 3 books and 4! permutations of the other 4 books, so the total number of allowed arrangements is 5 x 3!4! = 720.

(c) If two particular books must occupy the ends, there are 2! = 2 permutations of the two books and 5! = 120 permutations of the other 5 books, so the total number of allowed arrangements is 2!5! = 240.

How many five-letter words can be made from five letters if (a) all letters are different, (b) two letters are identical, (c) all letters are different but two particular letters cannot be adjacent?

[from Spiegel, Murray R. 1975,

SOLUTION

(a) If the five letters are all different, 5! = 120 five-letter words can be made from them.

(b) If two letters are identical, 5!/2! = 60 five-letter words can be made from them.

(c) Let A and B be the two letters that cannot be adjacent. Then the possible positions of A and B are

AXBXX

AXXBX

AXXXB

XAXBX

XAXXB

XXAXB

or the positions with B and A interchanged. This gives twelve possible positions for A and B. For each of these, there are 3! ways to arrange the remaining three letters, so the total number of five-letter words that can be made is 12 x 3! = 72.

Find the number of (a) combinations and (b) permutations of four letters each that can be made from the letters of the word "Tennessee."

[from Spiegel, Murray R. 1975,

SOLUTION

The word "Tennessee" contains one t, four e's, two n's, and two s's. We can solve this problem by considering different categories of possible combinations, finding all possible combinations in each category, and calculating the number of permutations for each combination. The table below shows ten categories, each defined by the number of e's and s's. The number of permutations of each combination is calculated using the fact that the number of permutations of n objects where there are n

Category | Combinations | Permutations |
---|---|---|

4 e's |
eeee |
1 |

3 e's |
eeeteeen eees |
4!/3! = 44!/3! = 4 4!/3! = 4 |

2 e's2 s's |
eess |
4!/(2!2!) = 6 |

2 e's1 s |
eesteesn |
4!/2! = 124!/2! = 12 |

2 e's0 s's |
eetneenn |
4!/2! = 124!/(2!2!) = 6 |

1 e2 s's |
esstessn |
4!/2! = 124!/2! = 12 |

1 e1 s |
estnesnn |
4! = 244!/2! = 12 |

1 e0 s's |
etnn |
4!/2! = 12 |

0 e's2 s's |
sstnssnn |
4!/2! = 124!/(2!2!) = 6 |

0 e's1 s |
stnn |
4!/2! = 12 |

Totals |
(a) 17 combinations |
(b) 163 permutations |

Find the number of (a) combinations and (b) permutations of four letters each that can be made from the letters in the word "Mississippi."

SOLUTION

The word "Mississippi" contains one m, four i's, four s's, and two p's. We can solve this problem by considering different categories of possible combinations, finding all possible combinations in each category, and calculating the number of permutations for each combination. The table below shows 13 categories, each defined by the number of i's and s's. We define the categories by specifying the number of i's and s's because these are the most frequently appearing letters in "Mississippi" and it is then easier to list all the possible combinations of the m and i's, if any, that fit into the category. The number of permutations of each combination is calculated using the fact that the number of permutations of n objects where there are n

Category | Combinations | Permutations |
---|---|---|

4 i's |
iiii |
1 |

3 i's |
iiimiiis iiip |
4!/3! = 44!/3! = 4 4!/3! = 4 |

2 i's2 s's |
iiss |
4!/(2!2!) = 6 |

2 i's1 s |
iismiisp |
4!/2! = 124!/2! = 12 |

2 i's0 s's |
iimpiipp |
4!/2! = 124!/(2!2!) = 6 |

1 i3 s's |
isss |
4!/3! = 4 |

1 i2 s's |
issmissp |
4!/2! = 124!/2! = 12 |

1 i1 s |
ismpispp |
4! = 244!/2! = 12 |

1 i0 s's |
impp |
4!/2! = 12 |

0 i's4 s's |
ssss |
1 |

0 i's3 s's |
sssmsssp |
4!/3! = 44!/3! = 4 |

0 i's2 s's |
ssmpsspp |
4!/2! = 124!/(2!2!) = 6 |

0 i's1 s |
smpp |
4!/2! = 12 |

Totals |
(a) 21 combinations |
(b) 176 permutations |

(a) A shelf contains 6 separate compartments. In how many ways can 4 indistinguishable marbles be placed in the compartments?

(b) Work the problem if there are n compartments and r marbles.

[from Spiegel, Murray R. 1975,

SOLUTION

(a) The number of ways n indistinguishable marbles can be placed in k compartments is

N(n, k) = (n+k-1)!/[n!(k-1)!]

The number of ways 4 indistinguishable marbles can be placed in 6 separate compartments is

N(4, 6) = (4+6-1)!/[4!(6-1)!] = 9!/(4!5!) = 126

(b) The number of ways r indistinguishable marbles can be placed in n compartments is

N(r, n) = (r+n-1)!/[r!(n-1)!]

(a) A shelf contains 6 separate compartments. In how many ways can 12 indistinguishable marbles be placed in the compartments so that no compartment is empty?

(b) Work the problem if there are n compartments and r marbles where r > n.

[from Spiegel, Murray R. 1975,

SOLUTION

(a) Each compartment must contain at least 1 marble. This uses up 6 of the marbles. The remaining 6 must be distributed among 6 compartments. The number of ways this can be done is

N(6, 6) = (6+6-1)!/[6!(6-1)!] = 11!/(6!5!) = 462

(b) Each compartment must contain at least 1 marble. This uses up n of the marbles. The remaining r-n must be distributed among n compartments. The number of ways this can be done is

N(r-n, n) = (r-n+n-1)!/[(r-n)!(n-1)!] = (r-1)!/[(r-n)!(n-1)!]

(a) How many gin rummy hands, consisting of 10 cards, can be dealt from an ordinary deck of 52 playing cards?

(b) How many bridge hands, consisting of 13 cards, can be dealt from an ordinary deck of 52 playing cards?

[from Freund, John E., and Simon, Gary A. 1997,

SOLUTION

(a) The number of gin rummy hands that can be dealt from an ordinary deck of 52 playing cards is the same as the number of combinations of 10 items picked from 52 items, which is

C(52, 10) = 52!/[10!(52-10)!] = 52!/(10!42!) = 1.582 x 10

(b) The number of bridge hands that can be dealt from an ordinary deck of 52 playing cards is the same as the number of combinations of 13 items picked from 52 items, which is

C(52, 13) = 52!/[13!(52-13)!] = 52!/(13!39!) = 6.350 x 10

The number of ways in which n distinct objects can be arranged in a circle is (n - 1)!.

(a) Present an argument to justify this formula.

(b) In how many ways can six persons be seated at a round table, if we care only who sits on whose left or right side?

(c) In how many ways can eight people form a circle for a folk dance?

(d) Four men and four women are forming a circle for a folk dance. In how many ways can this be done if we require that the men and women alternate positions?

[from Freund, John E., and Simon, Gary A. 1997,

SOLUTION

(a) The number of permutations of n distinguishable objects is n!. If the n objects are arranged in a circle, then for each particular permutation there are n - 1 additional permutations which are merely clockwise or counterclockwise shifts of all the objects. Thus, the number of ways n distinct objects can be arranged in a circle, if all we care about is the relative ordering, is n!/n = (n - 1)!.

(b) The number of ways six people can be seated at a round table if we only care about the relative ordering is (6 - 1)! = 5! = 120.

(c) The number of ways eight people can be seated at a round table if we only care about the relative ordering is (8 - 1)! = 7! = 5040.

(d) There are (4 - 1)! = 3! = 6 ways to arrange the four men in alternate positions in a circle. For each of these orderings, there are 4! = 24 ways to arrange the four women. We use 4! and not (4 - 1)! because, once the men have been assigned positions, we should consider all permutations of the women, even those which are merely clockwise or counterclockwise shifts of another, because these result in different overall orderings of the eight people. Thus, the number of ways we can arrange the four men and four women is 6 x 24 = 144.

In how many ways can 3 men and 3 woman be seated at a round table if (a) no restriction is imposed, (b) two particular women cannot sit together, (c) each woman is to be between two men?

[from Spiegel, Murray R. 1975,

SOLUTION

(a) The number of ways to arrange n objects among n positions in a row is n!. In the case of a circular arrangement, there are n permutations of a particular arrangement which are really the same ordering except that all objects are shifted clockwise or counterclockwise by the same amount. Thus, the number of ways to arrange n objects among n positions in a circle is n!/n = (n-1)!. The number of ways to arrange 6 people at a round table with 6 seats, if no restrictions are imposed, is (6-1)! = 5! = 120.

(b) If two particular women cannot sit together, we have to subtract all cases where the women would be seated together. If the seats are numbered 1 through 6, these are of the form

123456

AABBBB

BAABBB

BBAABB

BBBAAB

BBBBAA

ABBBBA

where the A's are the two women and the B's are the other people. Since we want to exclude permutations which are simply shifts of a particular ordering, we can consider only the permutations of the form

AABBBB

or any one of the other five permutations listed. There are 2!4! = 48 such permutations. Thus, the number of ways 3 men and 3 women can be seated at a round table, if two particular women cannot sit together, is 120 - 48 = 72.

(c) If each woman is to be between two men, there are 2! = 2 ways to seat the women, and 3! = 6 ways to seat the men, or a total of 2!3! = 12 ways to seat the 3 men and 3 women. We use 2! for the women because we want to exclude permutations of the women which are simply shifts of a given permutation. We use 3! for the men because once the women have been assigned seats, any permutation of the men, including those which are shifts of another permutation, are allowed, since these result in different overall arrangements of the 6 people which are not just shifts of one particular permutations.

C(N, k) = N!/[k!(N-k)!]

This is also the number of combinations of k items picked from N items if the items are all distinguishable. The probability of a particular combination of k successes and N - k failures is

p(N, k) = p

The total probability of having k successes and N - k failures is

P(N, k) = C(N, k)p(N, k) = {N!/[k!(N-k)!]}p

In this case, each "experiment" is a coin toss, "success" is a heads, "failure" is a tails, and the experiment is repeated three times. The probability of success is the probability of a heads, which is p = 0.5, and the probability of failure is the probability of a tails, which is q = 0.5. Since we want to know the probability of getting two heads in three tosses, N = 3 and k = 2. Using (1), we get

P(3, 2) = C(3, 2)p(3, 2) = {3!/[2!(3-2)!]}(0.5)

We can understand how this formula works as follows. First we determine the number of ways we can get two heads or the number of ways we can distribute two heads among three coin tosses. Call the two heads H1 and H2, and the single tails T. The two heads and one tails can be assigned to the coin tosses as follows:

H1 H2 T

H1 T H2

H2 H1 T

T H1 H2

H2 T H1

T H2 H1

It appears that there are 3! = 6 possible ways to do this. But, in reality, the two heads are indistinguishable. Switching H1 and H2 gives an equivalent result. If we replace the H1 and H2 with H, we see that the six results are:

H H T

H T H

H H T

T H H

H T H

T H H

The only distinguishable results with two heads are:

H H T

H T H

T H H

The number of distinguishable results is 6!/2! = 3. We divide the 6! by 2! because there are 2! permutations of a given assignment of two heads, but these are indistinguishable. The probability of each of the three distinguishable results is (0.5)

A fair coin is tossed five times what is the probability of getting a sequence of three heads?

[from Rice, John A. 1995,

SOLUTION

There are 2

Total Number of Heads | Number of Possible Outcomes | Possible Outcomes |
---|---|---|

0 |
1 |
TTTTT |

1 |
5 |
HTTTTTHTTT TTHTT TTTHT TTTTH |

2 |
10 |
HHTTTHTHTT HTTHT HTTTH THHTT THTHT THTTH TTHHT TTHTH TTTHH |

3 |
10 |
TTHHH*THTHH THHTH THHHT* HTTHH HTHTH HTHHT HHTTH HHTHT HHHTT* |

4 |
5 |
HHHHT*HHHTH* HHTHH HTHHH* THHHH* |

5 |
1 |
HHHHH* |

Eight of the possible outcomes contain a sequence of three heads, so the probability of obtaining a sequence of three heads is 8/32 = 1/4. Back to Main Menu

Suppose that you repeatedly shake six coins in your hand and drop them on the table. Construct a table showing the number of microstates that correspond to each macrostate. What is the probability of obtaining (a) three heads and three tails, and (b) six heads?

[from Giancoli, Douglas C. 1998,

SOLUTION

In each of the following cases, determine how many ways the indicated macrostate can be achieved. These are all combinations of heads and tails that correspond to the indicated macrostate.

This can be considered an experiment repeated 6 times where the probability of "success" (heads) is p = 1/2 and the probability of "failure" (tails) is q = 1 - p = 1/2. The number of ways to get k heads and 6 - k tails is (6 k) = 6!/k!(6-k)!.

Macrostates | Number of Microstates | Microstates |
---|---|---|

6 heads, 0 tails |
6!/[0!(6-0)!] = 1 |
HHHHHH |

5 heads, 1 tail |
6!/[5!(6-5)!] = 6 |
HHHHHTHHHHTH HHHTHH HHTHHH HTHHHH THHHHH |

4 heads, 2 tails |
6!/[4!(6-2)!] = 15 |
HHHHTTand 14 others |

3 heads, 3 tails |
6!/[3!(6-3)!] = 20 |
HHHTTTand 19 others |

2 heads, 4 tails |
6!/[2!(6-2)!] = 15 |
HHTTTTand 14 others |

1 head, 5 tails |
6!/[1!(6-1)!] = 6 |
TTTTTHTTTTHT TTTHTT TTHTTT THTTTT HTTTTT |

0 heads, 6 tails |
6!/[0!(6-0)!] |
TTTTTT |

The total number of possible microstates is 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64. This is as expected since each flipped coin can have two possible outcomes, and there are six coins, so the number of possible microstates is 2

(a) The probability of obtaining three heads and three tails is 20/64 = 5/16, which can also be obtained by evaluating {6!/[3!(6-3)!]}(1/2)

(b) The probability of obtaining six heads is 1/64, which can also be obtained by evaluating

{6!/[6!(6-6)!]}(1/2)

Which is more likely: 9 heads in 10 tosses of a fair coin or 18 heads in 20 tosses?

[from Rice, John A. 1995,

SOLUTION

The number of ways to get k heads in N coin tosses is the same as the number of combinations of k items picked from N items, which is

C(N, k) = N!/[(N-k)!k!]

The probability of each combination is

p(N, k) = a

where a is the probability of the outcome which occurs k times and b is the probability of the outcome which occurs N - k times. In the case of coin tosses, a = 1/2 is the probability of getting a heads in a single coin toss, and b = 1/2 is the probability of getting a tails in a single coin toss. The probability of getting k heads in N coin tosses is

P(N, k) = C(N, k)p(N, k) = {N!/[(N-k)!k!]}a

= {N!/[(N-k)!k!]}(1/2)

The probability of getting 9 heads in 10 tosses is

P(10, 9) = {10!/[(10-9)!9!]}(1/2)

The probability of getting 18 heads in 20 tosses is

P(20, 18) = {20!/[(20-18)!18!]}(1/2)

Since P(10, 9) > P(20, 18), 9 heads in 10 tosses is more likely.

Calculate the probabilities, when you throw two dice, of obtaining (a) a 5, (b) an 11, and (c) any other value.

[(a) and (b) from Giancoli, Douglas C. 1998,

SOLUTION

(a) There are six possible values on each die, so there are 6 x 6 = 36 possible combinations [36 microstates] on the two dice.

A five can be obtained with (4, 1), (3, 2), (2, 3), or (1, 4) [4 microstates], so the probability of obtaining a five is 4/36 = 1/9.

(b) An eleven can be obtained with (6, 5) or (5, 6) [2 microstates], so the probability of obtaining an eleven is 2/36 = 1/18.

(c) 2: (1, 1) [1 microstate] => 1/36

3: (2, 1), (1, 2) [2 microstates] => 2/36 = 1/18

4: (3, 1), (2, 2), (1, 3) [3 microstates] => 3/36 = 1/12

6: (5, 1), (4, 2), (3, 3), (2, 4), (1, 5) [5 microstates] => 5/36

7: (6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6) [6 microstates] => 6/36 = 1/6

8: (6, 2), (5, 3), (4, 4), (3, 5), (2, 6) [5 microstates] => 5/36

9: (6, 3), (5, 4), (4, 5), (3, 6) [4 microstates] => 4/36 = 1/9

10: (6, 4), (5, 5), (4, 6) [3 microstates] => 3/36 = 1/12

12: (6, 6) [1 microstate] => 1/36

The probability of obtaining a 2, 3, 4, 6, 7, 8, 9, 10, or 12 is 1/36 + 1/18 + 1/12 + 5/36 + 1/6 + 5/36 + 1/9 + 1/12 + 1/36 = 30/36 = 5/6. As expected, the sum of this and the probabilities obtained in parts (a) and (b) is 5/6 + 1/9 + 1/18 = 1.

A pair of dice is tossed repeatedly. Find the probability that an 11 occurs for the first time on the 6th toss.

[from Spiegel, Murray R. 1975,

SOLUTION

Each throw of the dice has 6 x 6 = 36 possible outcomes because there are 6 possible outcomes on each die. Since two of the possible outcomes, (6, 5) and (5, 6), result in an 11, the probability that each throw of the dice will result in an 11 is 2/36 = 1/18, and the probability that each throw of the dice will not result in an 11 is 34/36 = 17/18. The probability that an 11 occurs for the first time on the 6th toss is equal to the probability that an 11 does not occur on any of the first five tosses and does occur on the 6th toss. Thus, the probability that an 11 occurs for the first time on the 6th toss is (17/18)

A poker player has cards 2, 3, 4, 6, and 8. He wishes to discard the 8 and replace it by another card which he hopes will be a 5 (in which case he gets an "inside straight"). What is the probability that he will succeed assuming that the other three players together have (a) one 5, (b) two 5's, (c) three 5's, (d) no 5? (e) Can the problem be worked if the number of 5's in the other players' hands is unknown? Explain.

[from Spiegel, Murray R. 1975,

SOLUTION

(a) The number of cards that are not in any players' hands is 52 - (4)(5) = 52 - 20 = 32. If the other players together have one 5, there are three 5's among the 32 cards, so the probability that the player will pick a 5 is 3/32.

(b) If the other players together have two 5's, there are two 5's among the 32 cards, so the probability that the player will pick a 5 is 2/32 = 1/16.

(c) If the other players together have three 5's, there is one 5 among the 32 cards, so the probability that the player will pick a 5 is 1/32.

(d) If there are no 5's in the other players' hands, there are four among the 32 cards, so the probability that the player will pick a 5 is 4/32 = 1/8.

(e) If the number of 5's in the other players' hands is unknown, each of the 5's could be anywhere among the 47 cards that the player does not already have. The probability that the player gets a 5 if he picks one card, from the 32 that are not in any players' hands, is 4/47.

Rank the following five-card hands in order of increasing probability:

(a) four aces and a king

(b) six of hearts, eight of diamonds, queen of clubs, three of hearts, jack of spades

(c) two jacks, two queens, and an ace

(d) any hand having no two equal-value cards

Explain your ranking using microstates and macrostates.

[Giancoli, Douglas C. 1998,

SOLUTION

(a) The probability of obtaining four aces and a king is equal to the [number of ways of obtaining four aces and a king] divided by the [total number of possible five-card hands].

The [total number of possible five-card hands] is equal to the [number of ways of picking the first card (52)] times the [number of ways of picking the second card (51)]...times the [number of ways of picking the fifth card (48)] = 52 x 51 x 50 x 49 x 48 = 52!/47!.

The [number of ways of obtaining four aces and a king] is equal to four times the [number of ways of obtaining four aces and a particular king, such as the king of spades] = 4 x [number of permutations of four aces and a particular king] = 4 x 5!.

The probability of obtaining four aces and a king is 4 x 5! / (52!/47!) = 4 x 5! / (52 x 51 x 50 x 49 x 48) = 1.53908 x 10

(b) The probability of obtaining a six of hearts (6H), eight of diamonds (8D), queen of clubs (QC), three of hearts (3H), and jack of spades (JS) is equal to the [number of ways of obtaining 6H, 8D, QC, 3H, and JS] divided by the [total number of possible five-card hands].

The [number of ways of obtaining 6H, 8D, QC, 3H, and JS] is the [number of permutations of 6H, 8D, QC, 3H, and JS] = 5!.

The probability of obtaining 6H, 8D, QC, 3H, and JS is 5!/(52!/47!) = 5!/(52 x 51 x 50 x 49 x 48) = 3.84769 x 10

(c) The probability of obtaining two jacks, two queens, and an ace is equal to the [number of ways of obtaining two jacks, two queens, and an ace] divided by the [total number of possible five-card hands].

The [number of ways of obtaining two jacks, two queens, and an ace] is equal to the [number of ways of obtaining two jacks] times the [number of ways of obtaining two queens] times the [number of ways of obtaining one ace] times the [number of permutations of the five cards once they have been picked] = (4!/2!2!) x (4!/2!2!) x 4 x 5! = 6 x 6 x 4 x 120 = 17280.

The probability of obtaining two jacks, two queens, and an ace is equal to 17280/(52!/47!) = 17280/(52 x 51 x 50 x 49 x 48) = 5.54068 x 10

(d) The probability of picking any hand having no two equal-value cards is equal to the [number of hands having no two equal-value cards] divided by the [total number of possible five-card hands] .

The [number of hands having no two equal-value cards] is equal to the [number of hands having cards with all different values], which is equal to the [number of ways of picking the first card (52)] times the [number of ways of picking the second card (48)] times the [number of ways of picking the third card (44)] times the [number of ways of picking the fourth card (40)] times the [number of ways of picking the fifth card (36)] = 52 x 48 x 44 x 40 x 36.

The probability of picking any hand having no two equal-value cards is 52 x 48 x 44 x 40 x 36 / (52!/47!) = 48 x 44 x 40 x 36 / (51!/47!) = 48 x 44 x 40 x 36 / (51 x 50 x 49 x 48) = 0.507082833 = 1/1.972064394. The number of microstates corresponding to the required macrostate is 52 x 48 x 44 x 40 x 36 = 158,146,560.

P(b) < P(a) < P(c) < P(d)

Estimate the probability that a bridge player will be dealt (a) all four aces (among 13 cards), and (b) all 13 cards of one suit.

[from Giancoli, Douglas C. 1998,

SOLUTION

(a) The probability that a bridge player will be dealt all four aces is equal to the [number of ways a thirteen-card hand can contain all four aces] divided by the [total number of possible thirteen-card hands].

The [total number of possible thirteen-card hands] is equal to the [number of ways of picking the first card (52)] times the [number of ways of picking the second card (51)]...times the [number of ways of picking the thirteenth card (40)] = 52 x 51 x 50 x ... x 40 = 52!/39!.

The [number of ways a thirteen-card hand can contain all four aces] is equal to the [number of ways the four aces can be distributed among the thirteen cards] times the [number of ways the remaining nine cards can be selected from the non-ace cards].

The [number of ways the four aces can be distributed among the thirteen cards] is equal to the [number of ways the first ace can be assigned a position (13)] times the [number of ways the second ace can be assigned a position (12)] times the [number of ways the third ace can be assigned a position (11)] times the [number of ways the fourth ace can be assigned a position (10)] = 13 x 12 x 11 x 10 = 13!/9!.

The [number of ways the remaining nine cards casn be selected from the non-ace cards] is equal to the [number of ways the first non-ace card can be picked (48)] times the [number of ways the second non-ace card can be picked (47)]...times the [number of ways the ninth non-ace card can be picked (40)] = 48 x 47 x ... x 40 = 48!/39!.

The [number of ways a thirteen-card hand can contain all four aces] is (13!/9!) x (48!/39!). The probability that a bridge player will be dealt all four aces is (13!/9!) x (48!/39!)/(52!/39!) = 13 x 12 x 11 x 10 / (52 x 51 x 50 x 49) = 0.002641 = 1/378.6364.

(b) The probability that a bridge player will be dealt all 13 cards of one suit is equal to the [number of ways the thirteen-card hand can contain all 13 cards of one suit] divided by the [total number of possible thirteen-card hands].

The [number of ways the thirteen-card hand can contain all 13 cards of one suit] is equal to four times the [number of ways the thirteen-card hand can contain all 13 cards of one particular suit, such as hearts], which is four times the [number of ways the 13 cards of one particular suit can be distributed in a thirteen-card hand], which is four times the [number of ways the first card can be assigned a position in the thirteen-card hand (13)] times the [number of ways the second card can be assigned a position in the thirteen-card hand (12)]...times the [number of ways the thirteenth card can be assigned a position in the thirteen-card hand (1)] = 4 x 13!.

The probability that a bridge player will be dealt all 13 cards of one suit is 4 x 13! / (52!/39!) = 4 x 13! / (52 x 51 x ... x 40) = 6.29908 x 10

A radio station that plays classical music has a "by request" program each Saturday evening. The percentages of requests for composers on a particular night are as follows:

Composer | Percentage |
---|---|

Bach |
5% |

Beethoven |
26% |

Brahms |
9% |

Dvorak |
2% |

Mendelssohn |
3% |

Mozart |
21% |

Schubert |
12% |

Schumann |
7% |

Tchaikovsky |
14% |

Wagner |
1% |

(a) What is the probability that the request is for one of the three B's?

(b) What is the probability that the request is not for one of the two S's?

(c) Neither Bach nor Wagner wrote any symphonies. What is the probability that the request is for a composer who wrote at least one symphony?

(d) What is the probability that the request is for a piece by one of the main characters of the movie

[from Devore, Jay, and Peck, Roxy 1994,

SOLUTION

(a) The probability that the request is for Bach, Beethoven, or Brahms is 0.05 + 0.26 + 0.09 = 0.40.

(b) The probability that the request is for Schubert or Schumann is 0.12 + 0.07 = 0.19. The probability that the request is not for Schubert or Schumann is 1.00 - 0.19 = 0.81.

(c) The probability that the request is for one of the composers other than Bach or Wagner is 1.00 - 0.05 - 0.01 = 0.94.

(d) The movie

After mixing a deck of 52 cards very well, 5 cards are dealt out.

(a) Show that (disregarding the order in which the cards are dealt) there are 2,598,960 possible hands, of which only 1287 are hands consisting entirely of spades. What is the probability that a hand will consist entirely of spades? What is the probability that a hand will consist entirely of a single suit?

(b) Show that exactly 63,206 hands contain only spades and clubs, with both suits represented. What is the probability that a hand consists entirely of spades and clubs with both suits represented?

(c) Using the result of part (b), what is the probability that a hand contains cards from exactly two suits?

[from Devore, Jay, and Peck, Roxy 1994,

SOLUTION

(a) The number of five-card hands is equal to the number of combinations of five cards that can be picked from 52 cards, which is

C(52, 5) = 52!/[5!(52-5)!] = 52!/(5!47!) = (52)(51)(50)(49)(48)/120 = 2,598,960

The number of five-card hands consisting entirely of spades is equal to the number of combinations of five cards that can be picked from thirteen cards, which is

C(13, 5) = 13!/[5!(13-5)!] = 13!/(5!8!) = (13)(12)(11)(10)(9)/120 = 1287

The probability that a hand will consist entirely of spades is

P(all spades) = C(13, 5) / C(52, 5) = 1287/2,598,960 = 4.952 x 10

The probability that a hand will consist entirely of a single suit is

P(single suit) = P(all spades) + P(all clubs) + P(all diamonds) + P(all hearts) = 4P(all spades)

= (4)(1287)/2,598,960 = 1.981 x 10

(b) We need to determine the number of five-card hands which contain spades and clubs. The number of five-card hands which contain k spades and 5 - k clubs is equal to the number of ways we can pick k cards from 13 cards times the number of ways we can pick 5 - k cards from 13 cards or

N(k, 5-k) = C(13, k)C(13, 5-k) = {13!/[k!(13-k)!]}(13!/{(5-k)![13-(5-k)]!}) br>

The following table shows the number of five-card hands which contain k spades and 5 - k clubs for k = 1, 2, 3, 4.

k (k spades + [5-k] clubs) | N(k, 5-k) |
---|---|

1 (1 spades + 4 clubs) |
[13!/(1!12!)][13!/(4!9!)] = (13)(13)(12)(11)(10)/24 = 9295 |

2 (2 spades + 3 clubs) |
[13!/(2!11!)][13!/(3!10!)] = (13)(12)(13)(12)(11)/12 = 22308 |

3 (3 spades + 2 clubs) |
[13!/(3!10!)][13!/(2!11!)] = (13)(12)(11)(13)(12)/12 = 22308 |

4 (4 spades + 1 clubs) |
[13!/(4!9!)][13!/(1!12!)] = (13)(12)(11)(10)(13)/24 = 9295 |

Total |
63206 |

Thus, exactly 63,206 hands contain only spades and clubs, with both suits represented. The probability that a hand consists entirely of spades and clubs with both suits represented is

P(spades + clubs) = 63,206/2,598,960 = 2.432 x 10

(c) The probability that a hand contains cards from exactly two suits is

P(two suits) = P(spades + clubs) + P(spades + diamonds) + P(spades + hearts)

+ P(clubs + diamonds) + P(clubs + hearts) + P(diamonds + hearts)

= 6P(spades + clubs) = (6)(63,206)/2,598,960 = 0.1459 Back to Main Menu

A library has six Colin Dexter mysteries on the shelf, five by Tony Hillerman, and four by Arthur Upfield.

(a) How many ways are there to select one book by each author?

(b) How many ways are there to select three books from among the 15 without regard to author?

(c) If three books are selected in a completely random fashion, what is the probability that the three are by different authors?

[from Devore, Jay, and Peck, Roxy 1994,

SOLUTION

(a) There are (6, 5, 4) ways to choose a (Dexter, Hillerman, Upfield) book. The number of ways there are to select one book by each author is 6 x 5 x 4 = 120.

(b) The number of ways to select three books from among the 15 without regard to author is equal to the number of combinations of 3 items picked from 15, which is

C(15, 3) = 15!/[3!(15-3)!] = 15!/(3!12!) = 455

(c) The number of ways of selecting three books in a completely random fashion is equal to the number of combinations of 3 items picked from 15, which is 455. The probability that the three are by different authors is 120/455 = 24/91 = 0.2637.

You have identified eight courses that you are seriously considering taking next term.

(a) How many ways are there to select five from among the eight without regard to order?

(b) Two of these are statistics courses, and you have definitely decided to take them. How many ways are there now to select five courses to take?

(c) If you randomly select five from among the eight courses, what is the probability that both the statistics courses will be on your schedule?

[from Devore, Jay, and Peck, Roxy 1994,

SOLUTION

(a) The number of ways to select five courses from among the eight is equal to the number of combinations of five items picked from eight, which is

C(8, 5) = 8!/[5!(8-5)!] = 8!/(5!3!) = 56

(b) Two of the five selected courses must be the two statistics courses. The number of ways of selecting three more courses from the remaining six is

C(6, 3) = 6!/[3!(6-3)!] = 6!/(3!3!) = 20

(c) If you randomly select five from among the eight courses, the probability that both the statistics courses will be on your schedule is 20/56 = 5/14 = 0.3571.

Here is a famous problem that shows coincidences are often not as unlikely to occur as you might think.

(a) Suppose that there are three people in a room. Disregarding the possibility of a February 29th birthday, how many possible combinations of the three birthdays are there?

(b) How many ways are there for the three people to have different birthdays?

(c) If each individual is equally likely to be born on any one of the 365 days, what is the probability that (i) all three people have different birthdays, (ii) at least two have the same birthday?

(d) Answer the questions posed in part (c) for a group of ten people.

(e) How many people need to be in a room before the probability that at least two have the same birthday exceeds 0.5?

[from Devore, Jay, and Peck, Roxy 1994,

SOLUTION

(a) Each person's birthday could be one of 365 different possible dates. The number of possible combinations of the three birthdays is therefore 365

(b) The number of ways for the three people to have different birthdays is (365)(364)(363) = 48228180.

(c) The probability that all three people have different birthdays is 48228180 / 48627125 = 0.9918.

The probability that at least two people have the same birthday is 1 - 0.9918 = 0.008204.

(d) The number of possible combinations of ten people's birthdays is 365

The number of ways for ten people to have different birthdays is 365!/355! = (365)(364)(363)(362)(361)(360)(359)(358)(357)(356) = 3.706 x 10

The probability that all ten people have different birthdays is (3.706 x 10

The probability that at least two people have the same birthday is 1 - 0.8831 = 0.1169.

(e) The number of possible combinations of n people's birthdays is 365

The number of ways for n people to have different birthdays is 365!/(365-n)!.

The probability that all n people have different birthdays is [365!/(365-n)!] / 365

The probability that at least two people have the same birthday is

P = 1 - [365!/(365-n)!] / 365

=> 0.5 > [365!/(365-n)!] / 365

=> log 0.5 > log[365!/(365-n)!] - n log 365

=> log 0.5 > log 365 + log 364 + ... + log(365-n+1) - n log 365

=> log 0.5 > f(n)

=> - 0.30103 > f(n) (1)

where

f(n) = log 365 + log 364 + ... + log(365-n+1) - n log 365

= S

If we find the smallest value of n for which (1) is satisfied, this will be the smallest value of n for which the probability that at least two people have the same birthday is greater than 0.5 (i.e., it is more likely that at least two people have the same birthday than it is that all the people have different birthdays).

The file DevorePeck4.60.061231.xls lists values of f(n) and 1 - 10

n | f(n) | 1 - 10^{f(n)} |
---|---|---|

1 |
0 |
0 |

2 |
- 0.00119 |
0.00274 |

3 |
- 0.00358 |
0.008204 |

10 |
- 0.05401 |
0.116948 |

22 |
- 0.28042 |
0.475695 |

23 |
- 0.30741 |
0.507297 |

40 |
- 0.9635 |
0.891232 |

41 |
- 1.01391 |
0.903152 |

56 |
- 1.93302 |
0.988332 |

57 |
- 2.00535 |
0.990122 |

We see that if n = (23, 41, 57), the probability that at least two people have the same birthday is greater than (0.5, 0.9, 0.99). Back to Main Menu

(a) Find the probability and odds of rolling eight or less with a pair of balanced dice.

(b) If a researcher randomly selected 8 of 80 households to be included in a study, find the probability and odds that any particular household will be included.

(c) Find the probability and odds of getting at least three heads in six flips of a balanced coin.

[from Freund, John E., and Simon, Gary A. 1997,

SOLUTION

(a) There are 36 possible results of rolling two dice. The following table lists the combinations which give various total values on the two dice.

Total | Combinations | Probability |
---|---|---|

2 |
(1, 1) |
1/36 |

3 |
(2, 1) (1, 2) |
2/36 = 1/18 |

4 |
(3, 1) (2, 2) (1, 3) |
3/36 = 1/12 |

5 |
(4, 1) (3, 2) (2, 3) (1, 4) |
4/36 = 1/9 |

6 |
(5, 1) (4, 2) (3, 3) (2, 4) (1, 5) |
5/36 |

7 |
(6, 1) (5, 2) (4, 3) (3, 4) (2, 5) (1, 6) |
6/36 = 1/6 |

8 |
(6, 2) (5, 3) (4, 4) (3, 5) (2, 6) |
5/36 |

9 |
(6, 3) (5, 4) (4, 5) (3, 6) |
4/36 = 1/9 |

10 |
(6, 4) (5, 5) (4, 6) |
3/36 = 1/12 |

11 |
(6, 5) (5, 6) |
2/36 = 1/18 |

12 |
(6, 6) |
1/36 |

The probability of rolling an eight or less is 26/36 = 13/18. The odds of rolling an eight or less are (13/18) / (1 - 13/18) = 13/5 or 13 to 5.

(b) The probability that any particular household will be included in the study is 8/80 = 1/10. The odds that any particular household will be included in the study is (8/80) / (1 - 8/80) = 8/72 = 1/9 or 1 to 9.

(c) The number of ways to get k heads in n coin flips is the same as the number of combinations of k items chosen from n items, which is n!/[k!(n-k)!]. The following table shows the number of ways six coin flips can result in different numbers of heads.

Total Heads | Combinations | Probability |
---|---|---|

6 |
6!/[6!(6-6)!] = 1 |
1/64 |

5 |
6!/[5!(6-5)!] = 6 |
6/64 = 3/32 |

4 |
6!/[4!(6-4)!] = 15 |
15/64 |

3 |
6!/[3!(6-3)!] = 20 |
20/64 = 5/16 |

2 |
6!/[2!(6-2)!] = 15 |
15/64 |

1 |
6!/[1!(6-1)!] = 6 |
6/64 = 3/32 |

0 |
6!/[0!(6-0)!] = 1 |
1/64 |

The probability of getting at least three heads is 42/64 = 21/32. The odds of getting at least three heads are (21/32) / (1 - 21/32) = 21/11 or 21 to 11. Back to Main Menu

A hotel gets cars for its guests from three rental agencies, 20 percent from agency X, 40 percent from agency Y, and 40 percent from agency Z. If 14 percent of the cars from X, 4 percent from Y, and 8 percent from Z need tune-ups, what is the probability that

(a) a car needing a tune-up will be delivered to one of the guests?

(b) if a car needing a tune-up is delivered to one of the guests, it came from agency Z?

[from Freund, John E., and Simon, Gary A. 1997,

SOLUTION

(a) From the given probabilities, we can deduce the probabilities that a car will come from a particular agency and either need a tune-up or not need a tune-up. These are summarized in the following table.

Agency | Needs Tune-Up | Does Not Need Tune-Up | Total |
---|---|---|---|

X |
(0.14)(0.20) = 0.028 |
(0.86)(0.20) = 0.172 |
0.20 |

Y |
(0.04)(0.40) = 0.016 |
(0.96)(0.40) = 0.384 |
0.40 |

Z |
(0.08)(0.40) = 0.032 |
(0.92)(0.40) = 0.368 |
0.40 |

Total |
0.028 + 0.016 + 0.032 = 0.076 |
0.172 + 0.384 + 0.368 = 0.924 |
1.000 |

Thus, the probability that a car needing a tune-up will be delivered to one of the guests is 0.076.

(b) If a car needing a tune-up is delivered to one of the guests, the probability that it came from agency Z is 0.032 / 0.076 = 0.4211. Back to Main Menu

If the probability is 0.26 that any one woman will name yellow or orange as her favorite color, what is the probability that four women, selected at random, will all name yellow or orange as their favorite color?

[from Freund, John E., and Simon, Gary A. 1997,

SOLUTION

The probability that all four women will name yellow or orange as their favorite color is 0.26

A library received 40 new books, including 12 historical novels. If four of these books are selected at random, what is the probability that not one of them is a historical novel?

[from Freund, John E., and Simon, Gary A. 1997,

SOLUTION

The probability that none of the four selected books is a historical novel is

P(0) = P

where P

P

The number of historical novels which could be selected stays at 12 for i = 1, 2, 3, and 4, but the total number of books available to be selected from keeps decreasing every time one is selected. We have the following values of P

i | P_{i} = 1 - k/[n-(i-1)] |
---|---|

1 |
1 - 12/40 = 0.7 |

2 |
1 - 12/39 = 0.692308 |

3 |
1 - 12/38 = 0.684211 |

4 |
1 - 12/37 = 0.675676 |

Thus,

P(0) = P

A movie producer feels that the odds are 8 to 1 that his new movie will not be rated G, 15 to 3 that it will not be rated PG, and 13 to 5 that it will not get either of these two ratings. Are the corresponding probabilities consistent?

[from Freund, John E., and Simon, Gary A. 1997,

SOLUTION

We have the following odds and probabilities:

Rating | Odds | Probability |
---|---|---|

not G |
8 to 1 |
8 / (8 + 1) = 8/9 |

not PG |
15 to 3 |
15 / (15 + 3) = 15/18 = 5/6 |

G |
(1/9) / (1 - 1/9) = 1/8 or 1 to 8 |
1 - 8/9 = 1/9 |

PG |
(1/6) / (1 - 1/6) = 1/5 or 1 to 5 |
1 - 5/6 = 1/6 |

G or PG |
(5/18) / (1 - 5/18) = 5/13 or 5 to 13 |
1/9 + 1/6 = 2/18 + 3/18 = 5/18 |

not G or PG |
(13/18) / (1 - 13/18) = 13/5 or 13 to 5 |
1 - 5/18 = 13/18 |

Thus, the probabilities corresponding to the stated odds are consistent. Back to Main Menu

The probability that George will get an M.A. degree is 0.40, and the probability that with an M.A. degree he will get a well-paying job is 0.85. What is the probability that he will get an M.A. degree and a well-paying job?

[from Freund, John E., and Simon, Gary A. 1997,

SOLUTION

The probability that George will get an M.A. degree and a well-paying job is (0.40)(0.85) = 0.34.

Gets Well-Paying Job | Does Not Get Well-Paying Job | Total | |
---|---|---|---|

Gets M.A |
(0.40)(0.85) = 0.34 |
(0.40)(0.15) = 0.06 |
0.40 |

Does Not Get M.A. |
- |
- |
0.60 |

Lie detectors have been used during wartime to uncover security risks. As is well known, lie detectors are not infallible. Let us suppose that the probability is 0.10 that the lie detector will fail to detect a person who is a security risk and that the probability is 0.08 that the lie detector will incorrectly label a person who is not a security risk. If 2 percent of the persons who are given the test are actually security risks, what is the probability that

(a) a person labeled a security risk by a lie detector test is in fact a security risk?

(b) a person who is cleared by a lie detector test is in fact not a security risk?

[from Freund, John E., and Simon, Gary A. 1997,

SOLUTION

Based on the information given above, we have the following probabilities:

Labeled Security Risk | Cleared | Total | |
---|---|---|---|

Security Risk |
(0.90)(0.02) = 0.018 |
(0.10)(0.02) = 0.002 |
0.02 |

Non-Security Risk |
(0.08)(0.98) = 0.0784 |
(0.92)(0.98) = 0.9016 |
0.98 |

Total |
0.018 + 0.0784 = 0.0964 |
0.002 + 0.9016 = 0.9036 |
1.00 |

(a) The probability that a person labeled a security risk by a lie detector test is in fact a security risk is 0.018 / 0.0964 = 0.1867.

(b) The probability that a person who is cleared by a lie detector test is in fact not a security risk is 0.9016 / 0.9036 = 0.9978. Back to Main Menu

A lot of n items contains k defectives, and m are selected randomly and inspected. How should the value of m be chosen so that the probability that at least one defective item turns up is 0.90? Apply your answer to (a) n = 1000, k = 10, and (b) n = 10,000, k = 100.

[from Rice, John A. 1995,

SOLUTION

(a) We need to find m so that the probability that no defective item turns up is less than 0.10. The probability that no defective item turns up if m samples are selected is

P(0) = (1 - k/n)[1 - k/(n-1)][1 - k/(n-2)]...{1 - k/[n-(m-1)]} = P

where

P

The term k/[n-(i-1)] is the probability that when the ith sample is picked, after the first i - 1 samples picked have all resulted in a nondefective, the ith sample is defective.

The term {1 - k/[n-(i-1)]} is the probability that when the ith sample is picked, after the first i - 1 samples picked have all resulted in a nondefective, the ith sample is also nondefective.

If n = 1000 and k = 10, the probability that the first sample (i = 1) picked is defective is k/n = 10/1000 = 0.01. The probability that the first sample picked is nondefective is

P

If the first sample picked is nondefective, the probability that the second sample (i = 2) picked is defective is k/(n-1) = 10/999. The probability that the second sample picked is nondefective is

P

If we calculate P

(b) If n = 10,000 and k = 100, the probability that the first sample (i = 1) picked is defective is k/n = 100/10,000 = 0.01. The probability that the first sample picked is nondefective is

P

If the first sample picked is nondefective, the probability that the second sample (i = 2) picked is defective is k/(n-1) = 100/9999. The probability that the second sample picked is nondefective is

P

If we calculate P

A multiple-choice test consists of 20 items, each with four choices. A student is able to eliminate one of the choices on each question as incorrect and chooses randomly from the remaining three choices. A passing grade is 12 items or more correct.

(a) What is the probability that the student passes?

(b) Answer the question in part (a) again, assuming that the student can eliminate two of the choices on each question.

[from Rice, John A. 1995,

SOLUTION

(a) Define

N = 20 = total number of test questions

a = 1/3 = probability of getting a particular question correct

b = 2/3 = probability of getting a particular question wrong

C(k) = number of possible ways (combinations) to get k questions correct and N - k questions wrong

p(k) = probability of each of the C(k) combinations that has k questions correct and N - k questions wrong

P(k) = total probability of getting k questions correct and N - k questions wrong

The number possible ways to get k questions correct and N - k questions wrong is the same as the number of combinations of k items selected from N items, which is

C(k) = N!/[(N-k)!k!]

The probability of each of these combinations is

p(k) = a

The total probability of getting k questions correct and N - k questions wrong is

P(k) = C(k)p(k) = {N!/[(N-k)!k!]}a

The probability of getting at least 12 questions correct and passing is

P(k

The calculation is done in the Excel sheet Rice2.13.061202.xls. The result is P(k

(b) The solution to this part of the problem is the same as for part (a) except that here a = 1/2 = b. The calculation is also shown in the Excel sheet, and the result is P(k

The World Series is a "best of seven" series in which the first baseball team that wins four games wins the series. Assuming that the two baseball teams, A and B, are equally good, find the probability that A will eventually win the World Series if

(a) it loses the first game.

(b) it loses the first two games.

(c) it loses the first three games.

SOLUTION

(a) Consider all possible outcomes of the World Series,

LWWWLLW

LWWLLWW

LWLLWWW

LLLWWWW

LLWLWWW

LLWWLWW

LLWWWLW

LWLWLWW

LWLWWLW

Each of these has a probability of (1/2)

LLWWWW

LWLWWW

LWWLWW

LWWWLW

Each of these has a probability of (1/2)

LWWWW

The probability that team A will win in five games is (1/2)

LLLL

The probability that team A will lose in four games is (1/2)

LWLLL

LLWLL

LLLWL

Each of these has a probability of (1/2)

LLLWWL

LLWLWL

LLWWLL

LWLLWL

LWLWLL

LWWLLL

Each of these has a probability of (1/2)

LLLWWWL

LWLLWWL

LWWLLWL

LWWWLLL

LLWLWWL

LLWWLWL

LLWWWLL

LWLWLWL

LWLWWLL

LWWLWLL

Each of these has a probability of (1/2)

The probability that team A will win the World Series is 5/32 + 1/8 + 1/16 = 11/32. The probability that team A will lose the World Series is 1/8 + 3/16 + 3/16 + 5/32 = 21/32.

(b) Consider all possible outcomes of the World Series,

LLLWWWW

LLWLWWW

LLWWLWW

LLWWWLW

Each of these has a probability of (1/2)

LLWWWW

The probability that team A will win in six games is (1/2)

LLLL

The probability that team A will lose in four games is (1/2)

Team A loses in five games. The fifth game must be one of its losses. The possible outcomes are of the form LLXXXL where "XXX" must have one W and one L. There are two ways of doing this:

LLLWL

LLWLL

Each of these has a probability of (1/2)

LLLWWL

LLWLWL

LLWWLL

Each of these has a probability of (1/2)

LLLWWWL

LLWLWWL

LLWWLWL

LLWWWLL

Each of these has a probability of (1/2)

The probability that team A will win the World Series is 1/8 + 1/16 = 3/16. The probability that team A will lose the World Series is 1/4 + 1/4 + 3/16 + 1/8 = 13/16.

(c) Consider all possible outcomes of the World Series,

LLLWWWW

The probability that team A will win in seven games is (1/2)

LLLL

The probability that team A will lose in four games is 1/2.

LLLWL

The probability that team A will lose in five games is (1/2)

LLLWWL

The probability that team A will lose in six games is (1/2)

LLLWWWL

The probability that team A will lose in seven games is (1/2)

The probability that team A will win the World Series is 1/16. The probability that team A will lose the World Series is 1/2 + 1/4 + 1/8 + 1/16 = 15/16.

Two teams, A and B, play a series of games. If team A has probability 0.4 of winning each game, is it to its advantage to play the best three out of five games or the best four out of seven? Assume the outcomes of successive games are independent.

[from Rice, John A. 1995,

SOLUTION

(a) The possible outcomes of a best of five series fall into six categories. Listed in best-to-worst order for team A, these are: Team A wins in 3, 4, or 5 games, or team A loses in 5, 4, or 3 games.

(i) If team A wins in N + 1 games, the (N+1)th game must have a win for team A and the first N games must have two wins and N - 2 losses.

The number of ways the first N games can have two wins for team A is the same as the number of combinations of two items picked from N items, which is

C(N, 2) = N!/[(N-2)!2!]

The probability of each of these combinations, which would mean that team A has a total of 3 wins and N - 2 losses, is

p(3, N-2) = a

where (a, b) = (0.4, 0.6) is the probability that team A (wins, loses) a particular game. The probability that team A wins in N + 1 games is

P

(ii) If team A loses in N + 1 games, the (N+1)th game must have a loss for team A and the first N games must have two losses and N - 2 wins.

The number of ways the first N games can have two losses for team A is the same as the number of combinations of two items picked from N items, which is

C(N, 2) = N!/[(N-2)!2!]

The probability of each of these combinations, which would mean that team A has a total of 3 losses and N - 2 wins, is

p(N-2, 3) = a

where (a, b) = (0.4, 0.6) is the probability that team A (wins, loses) a particular game. The probability that team A loses in N + 1 games is

P

P

The following table shows the possible outcomes of a best of five series.

Overall Outcome | Number of Ways This Can Happen | Probability of Each Outcome | Total Probabiliity |
---|---|---|---|

Team A Wins in 3 GamesWWW |
1 |
(0.4)^{3} = 0.064 |
0.064 |

Team A Wins in 4 GamesXXXW XXX = 2W + 1L |
3!/[(3-2)!2!] = 3 |
(0.4)^{3}(0.6) = 0.0384 |
0.1152 |

Team A Wins in 5 GamesXXXXW XXXX = 2W + 2L |
4!/[(4-2)!2!] = 6 |
(0.4)^{3}(0.6)^{2} = 0.02304 |
0.13824 |

Team A Loses in 5 GamesXXXXL XXXX = 2L + 2W |
4!/[(4-2)!2!] = 6 |
(0.4)^{2}(0.6)^{3} = 0.03456 |
0.20736 |

Team A Loses in 4 GamesXXXL XXX = 2L + 1W |
3!/[(3-2)!2!] = 3 |
(0.4)(0.6)^{3} = 0.0864 |
0.2592 |

Team A Loses in 3 GamesLLL |
1 |
(0.6)^{3} = 0.216 |
0.216 |

The total probability that team A will win a best of five series is 0.064 + 0.1152 + 0.13824 = 0.31744.

(b) The possible outcomes of a best of seven series fall into eight categories. Listed in best-to-worst order for team A, these are: Team A wins in 4, 5, 6, or 7 games, or team A loses in 7, 6, 5, or 4 games.

(i) If team A wins in N + 1 games, the (N+1)th game must have a win for team A and the first N games must have three wins and N - 3 losses.

The number of ways the first N games can have three wins for team A is the same as the number of combinations of three items picked from N items, which is

C(N, 3) = N!/[(N-3)!3!]

The probability of each of these combinations, which would mean that team A has a total of 4 wins and N - 3 losses, is

p(4, N-3) = a

where (a, b) = (0.4, 0.6) is the probability that team A (wins, loses) a particular game. The probability that team A wins in N + 1 games is

P

(ii) If team A loses in N + 1 games, the (N+1)th game must have a loss for team A and the first N games must have three losses and N - 3 wins.

The number of ways the first N games can have three losses for team A is the same as the number of combinations of three items picked from N items, which is

C(N, 3) = N!/[(N-3)!3!]

The probability of each of these combinations, which would mean that team A has a total of 4 losses and N - 3 wins, is

p(N-3, 4) = a

where (a, b) = (0.4, 0.6) is the probability that team A (wins, loses) a particular game. The probability that team A loses in N + 1 games is

P

P

The following table shows the possible outcomes of a best of seven series.

Overall Outcome | Number of Ways This Can Happen | Probability of Each Outcome | Total Probabiliity |
---|---|---|---|

Team A Wins in 4 GamesWWWW |
1 |
(0.4)^{4} = 0.0256 |
0.0256 |

Team A Wins in 5 GamesXXXXW XXXX = 3W + 1L |
4!/[(4-3)!3!] = 4 |
(0.4)^{4}(0.6) = 0.01536 |
0.06144 |

Team A Wins in 6 GamesXXXXXW XXXXX = 3W + 2L |
5!/[(5-3)!3!] = 10 |
(0.4)^{4}(0.6)^{2} = 0.009216 |
0.09216 |

Team A Wins in 7 GamesXXXXXXW XXXXXX = 3W + 3L |
6!/[(6-3)!3!] = 20 |
(0.4)^{4}(0.6)^{3} = 0.0055296 |
0.110592 |

Team A Loses in 7 GamesXXXXXXL XXXXXX = 3L + 3W |
6!/[(6-3)!3!] = 20 |
(0.4)^{3}(0.6)^{4} = 0.0082944 |
0.165888 |

Team A Loses in 6 GamesXXXXXL XXXXX = 3L + 2W |
5!/[(5-3)!3!] = 10 |
(0.4)^{2}(0.6)^{4} = 0.020736 |
0.20736 |

Team A Loses in 5 GamesXXXXL XXXX = 3L + 1W |
4!/[(4-3)!3!] = 4 |
(0.4)(0.6)^{4} = 0.05184 |
0.20736 |

Team A Loses in 4 GamesLLLL |
1 |
(0.6)^{4}) = 0.1296 |
0.1296 |

The total probability that team A will win a best of seven series is 0.0256 + 0.06144 + 0.09216 + 0.110592 = 0.289792. Team A is more likely to win a best of five series. Back to Main Menu

y(x) = ax + b

where a and b are constants, by minimizing the root-mean-square (RMS) differences between the values of y(x

f = S

The partial derivatives of f with respect to a and b are

∂f/∂a = S

and

∂f/∂b = S

= 2(a S

If we define

S

S

S

S

we can write

∂f/∂a = 2(aS

∂f/∂b = 2(aS

Next, we set the partial derivatives equal to zero and solve for the values of a and b which minimize f.

∂f/∂a = 2(aS

∂f/∂b = 2(aS

Substituting (2) into (1), we get

aS

=> NaS

Substituting (3) into (2), we get

b = (1/N)S

= [(1/N)S

= [S

= (S

The following x and y values, in cm, are measured:

x (cm) | y (cm) |
---|---|

-4.375 |
0.000 |

-3.828 |
0.250 |

-3.281 |
1.250 |

-2.734 |
1.500 |

-2.188 |
1.750 |

-1.641 |
1.750 |

-1.094 |
2.000 |

-0.547 |
2.125 |

0.000 |
2.250 |

0.547 |
2.125 |

1.094 |
2.000 |

1.641 |
1.750 |

2.188 |
1.750 |

2.734 |
1.500 |

3.281 |
1.250 |

3.828 |
0.250 |

4.375 |
0.000 |

Find the function of the form

y = B sqrt[1 - (x/A)

where A, B, and C are constants, which best fits the data.

SOLUTION

To solve this problem, we write a Fortran program which constructs a three-dimensional grid of points (A, B, C) which represent combinations of A, B, and C to test. For each set (A, B, C), calculate the value of y which corresponds to each value of x in the above table, and the root-mean-square deviation from the data,

σ = {(1/N) S

where N = 17 is the number of data points,

y(x

and (x

The program takes as input the minimum and maximum values of A, B, and C, which define the range of values to be tested, and the number of grid points in each direction (A, B, C). When the grid point yielding the lowest value of σ is found, a new grid is constructed surrounding this grid point with the grid cell size reduced by a factor of 0.95. The grid point in this new grid which minimizes σ is found. The process is repeated until the optimal values of A, B, and C converge.

An example of a program which does this is bestfit060318.f. If the minimum and maximum values of A, B, and C are all chosen to be 0 and 10, the program converges to (A, B, C) = (5.535, 5.785, 3.626) with σ = 0.1482. A graph of the results can be seen in Bestfit060312.xls. Back to Main Menu