Math 461, Solution to Written Homework 4

[Pages:3]Math 461, Solution to Written Homework 4

1. Two teams play a series of games; the first team to win 4 games is declared the overall

winner.

Suppose

that

the

two

teams

are

evenly

matched

and

each

has

probability

1 2

winning

each game, independent of the outcomes of other games. Find the expected number of games

played.

Solution. Let X be the number of games played. Then P(X = 4) = 2?2-4 = 2-3, P(X =

5) = 2

4 1

? 2-5 = 2-2, P(X = 6) = 2

5 2

? 2-6

=

5 16

,

P(X

=

7)

=

2

6 3

?

2-7

=

5 16

.

Thus

E (X )

=

4

?

1 8

+

5

?

1 4

+

6

?

5 16

+

7

?

5 16

.

2. There are n types of coupons. Each time a coupon is collected, it is equally likely to

be any of the n types. Suppose that k coupons are to be collected. Let Ai be the event that

there is at least one type i coupon among those collected. Find (a) P(A1), (b) P(A1 A2).

Solution.

P(A1) = 1 -

n-1 n

k

.

P(A1 A2) = 1 -

n-2 n

k

.

3. On a multiple choice exam with 4 possible answers for each of the 5 questions, what is the probability that a student would get 4 or more correct answers just by guessing?

Solution Let E4 be the event that the student gets exactly 4 correct and E5 the event that the student gets all 5 correct. Then the desired probability is

P(E4)

+

P(E5)

=

5(

1 4

)4

3 4

+

(

1 4

)5.

4. When coin 1 is flipped, it lands on heads with probability .4; when coin 2 is flipped, it lands on heads with probability .7. One of these coins is randomly chosen and flipped 10 times. (a) what is the probability that exactly 7 of the 10 flips lands on heads? (b) Given that the first of these 10 flips lands heads, what is the conditional probability that exactly 7 of these 10 flips lands on heads?

1

Solution Let E1 be the event that coin 1 is selected and E2 the event that coin 2 is selected. Let F be the event that exactly 7 of the 10 flips lands on heads, let G be the event

that the fist flip is heads. Then

P(F )

=

P(E1)P(F |E1)

+

P(E2)P(F |E2)

=

1 2

10 7

(.4)7(.6)3

+

1 2

10 7

(.7)7(.3)3

and

P(F |G)

=

P(F G) P(G)

=

P(E1)P(F G|E1) P(E1)P(G|E1)

+ +

P(E2)P(F G|E2) P(E2)P(G|E2)

1 2

(.4)

9 6

(.4)6(.6)3

+

1 2

(.7)

9 6

(.7)6(.3)3

=

1 2

(.4)

+

1 2

(.7)

.

5. The monthly worldwide average number of airplane crashes of commercial airlines is 3.5. What is the approximate probability that there will be (a) at least 2 such accidents in the next month; (b) at most one such accident in the next month?

Solution The number X of airplane crashes in the next month is approximately a Poisson random variable with parameter 3.5. Thus the probability that there will be at least 2 such accidents in the next month is

1 - P(X = 0) - P(X = 1) = 1 - e-3.5 - 3.5e-3.5.

The probability that there will be at most one such accident in the next month is P(X = 0) + P(X = 1) = e-3.5 + 3.5e-3.5.

6. A certain typing agency has 2 typists. The average number of errors per article is 3 when typed by the first typist and 4.2 when typed by the second typist. If your article is equally likely to be typed by either typist, approximate the probability that it will have no errors.

Solution The number X of typos is approximately a Poisson random variable with parameter 3 when typed by the first typist, and is Poisson random variable with parameter 4.2 when typed by the second typist. Let E1 be the event that the article is typed by the first typist, and E2 the event that the article is typed by the second typist. Then

P(X

=

0)

=

P(E1)P(X

=

0|E1)

+ P(E2)P(X

=

0|E2)

=

1 2

e-3

+

1 2

e-4.2.

7. A fair die is rolled until a 6 appears. Find the probability that at most 6 rolls are needed.

Solution The number X of rolls needed to get a 6 is a geometric random variable with

parameter

1 6

.

Thus

P(X

6)

=

1

-

P(X

7)

=

1

-

(

5 6

)6.

2

8. Suppose that independent trials, each results in a success with probability p (0, 1),

are performed until a total of k successes are accumulated. Let X be the number of trials

needed. Find the probability mass function of X.

Solution To get total of k successes, one needs at least k trials. Let X be number of

trials needed to get a total of k successes. In order for X be equal to n, the n-th trial must

be a success and there must be exactly k - 1 successes in the first n - 1 trials. Thus the probability mass function of X is

n - 1 p(x) = 0 k - 1

pk(1 - p)n-k x = k, k + 1, . . . otherwise.

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