Some Discrete Probability

Chapter 5 Some Discrete Probability

Distributions

5.1 This is a uniform distribution: f (x) = 1

10 , for x = 1, 2, . . . , 10.

Therefore P (X < 4) =

f (x) = 3 10 .

x=1

5.2 Binomial distribution with n = 12 and p = 0.5. Hence P (X = 3) = P (X ≤ 3) − P (X ≤ 2) = 0.0730 − 0.0193 = 0.0537.

5.4 For n = 5 and p = 3/4, we have (a) P (X = 2) = 5

(b) P (X ≤ 3) =

b(x; 5, 3/4) = 1 − P (X = 4) − P (X = 5)

5.5 We are considering a b(x; 20, 0.3). (a) P (X ≥ 10) = 1 − P (X ≤ 9) = 1 − 0.9520 = 0.0480.

(b) P (X ≤ 4) = 0.2375. (c) P (X = 5) = 0.1789. This probability is not very small so this is not a rare event.

Therefore, P = 0.30 is reasonable.

5.6 For n = 6 and p = 1/2. (a) P (2 ≤ X ≤ 5) = P (X ≤ 5) − P (X ≤ 1) = 0.9844 − 0.1094.

(b) P (X < 3) = P (X ≤ 2) = 0.3438.

5.7 p = 0.7.

60 Chapter 5 Some Discrete Probability Distributions (a) For n = 10, P (X < 5) = P (X ≤ 4) = 0.0474.

(b) For n = 20, P (X < 10) = P (X ≤ 9) = 0.0171.

5.8 For n = 8 and p = 0.6, we have

(a) P (X = 3) = b(3; 8, 0.6) = P (X ≤ 3) − P (X ≤ 2) = 0.1737 − 0.0498 = 0.1239. (b) P (X ≥ 5) = 1 − P (X ≤ 4) = 1 − 0.4059 = 0.5941.

5.9 For n = 15 and p = 0.25, we have (a) P (3 ≤ X ≤ 6) = P (X ≤ 6) − P (X ≤ 2) = 0.9434 − 0.2361 = 0.7073.

(b) P (X < 4) = P (X ≤ 3) = 0.4613. (c) P (X > 5) = 1 − P (X ≤ 5) = 1 − 0.8516 = 0.1484.

5.10 From Table A.1 with n = 12 and p = 0.7, we have (a) P (7 ≤ X ≤ 9) = P (X ≤ 9) − P (X ≤ 6) = 0.7472 − 0.1178 = 0.6294.

(b) P (X ≤ 5) = 0.0386. (c) P (X ≥ 8) = 1 − P (X ≤ 7) = 1 − 0.2763 = 0.7237.

5.11 From Table A.1 with n = 7 and p = 0.9, we have P (X = 5) = P (X ≤ 5) − P (X ≤ 4) = 0.1497 − 0.0257 = 0.1240.

5.12 From Table A.1 with n = 9 and p = 0.25, we have P (X < 4) = 0.8343.

5.13 From Table A.1 with n = 5 and p = 0.7, we have P (X ≥ 3) = 1 − P (X ≤ 2) = 1 − 0.1631 = 0.8369.

5.14 (a) n = 4, P (X = 4) = 1 − 0.3439 = 0.6561. (b) Assuming the series went to the seventh game, the probability that the Bulls won

3 of the first 6 games and then the seventh game is given by

(c) The probability that the Bulls win is always 0.9.

5.15 p = 0.4 and n = 5. (a) P (X = 0) = 0.0778.

(b) P (X < 2) = P (X ≤ 1) = 0.3370. (c) P (X > 3) = 1 − P (X ≤ 3) = 1 − 0.9130 = 0.0870.

Solutions for Exercises in Chapter 5

5.16 Probability of 2 or more of 4 engines operating when p = 0.6 is

P (X ≥ 2) = 1 − P (X ≤ 1) = 0.8208,

and the probability of 1 or more of 2 engines operating when p = 0.6 is

P (X ≥ 1) = 1 − P (X = 0) = 0.8400.

The 2-engine plane has a slightly higher probability for a successful flight when p = 0.6.

5.17 Since µ = np = (5)(0.7) = 3.5 and σ 2 = npq = (5)(0.7)(0.3) = 1.05 with σ = 1.025. Then µ ± 2σ = 3.5 ± (2)(1.025) = 3.5 ± 2.050 or from 1.45 to 5.55. Therefore, at least 3/4 of the time when 5 people are selected at random, anywhere from 2 to 5 are of the

opinion that tranquilizers do not cure but only cover up the real problem.

5.18 (a) µ = np = (15)(0.25) = 3.75.

(b) With k = 2 and σ = √npq = (15)(0.25)(0.75) = 1.677, µ ± 2σ = 3.75 ± 3.354 or from 0.396 to 7.104.

5.19 Let X 1 = number of times encountered green light with P (Green) = 0.35,

X 2 = number of times encountered yellow light with P (Yellow) = 0.05, and

X 3 = number of times encountered red light with P (Red) = 0.60. Then

f (x x ,x ,x )= (0.35) 1 x

5.21 Using the multinomial distribution with required probability is

5.22 Using the multinomial distribution, we have 8 5,2,1 (1/2) 5 (1/4) 2 (1/4) = 21/256.

5.23 Using the multinomial distribution, we have

5.24 p = 0.40 and n = 6, so P (X = 4) = P (X ≤ 4)−P (X ≤ 3) = 0.9590−0.8208 = 0.1382.

5.25 n = 20 and the probability of a defective is p = 0.10. So, P (X ≤ 3) = 0.8670.

62 Chapter 5 Some Discrete Probability Distributions

5.26 n = 8 and p = 0.60; (a) P (X = 6) = 8

(b) P (X = 6) = P (X ≤ 6) − P (X ≤ 5) = 0.8936 − 0.6846 = 0.2090.

5.27 n = 20 and p = 0.90; (a) P (X = 18) = P (X ≤ 18) − P (X ≤ 17) = 0.6083 − 0.3231 = 0.2852.

(b) P (X ≥ 15) = 1 − P (X ≤ 14) = 1 − 0.0113 = 0.9887. (c) P (X ≤ 18) = 0.6083.

5.28 n = 20; (a) p = 0.20, P (X ≥ x) ≤ 0.5 and P (X < x) > 0.5 yields x = 4.

(b) p = 0.80, P (Y ≥ y) ≥ 0.8 and P (Y < y) < 0.2 yields y = 14.

5.29 Using the hypergeometric distribution, we get

5.30 P (X ≥ 1) = 1 − P (X = 0) = 1 − h(0; 15, 3, 6) = 1 − 53

5.31 Using the hypergeometric distribution, we get h(2; 9, 6, 4) = 9 = 5 .

5.32 (a) Probability that all 4 fire = h(4; 10, 4, 7) = 1

(b) Probability that at most 2 will not fire = h(x; 10, 4, 3) = 29 30 .

x=0

5.33 h(x; 6, 3, 4) = 3−x )

4 ( 2 x )(

6 , for x = 1, 2, 3.

( 3 ) P (2 ≤ X ≤ 3) = h(2; 6, 3, 4) + h(3; 6, 3, 4) = 4

5.35 P (X ≤ 2) = h(x; 50, 5, 10) = 0.9517.

x=0

5.36 (a) P (X = 0) = h(0; 25, 3, 3) = 77

(b) P (X = 1) = h(1; 25, 3, 1) = 3 25 .

5.37 (a) P (X = 0) = b(0; 3, 3/25) = 0.6815.

Solutions for Exercises in Chapter 5

(b) P (1 ≤ X ≤ 3) =

b(x; 3, 1/25) = 0.1153.

x=1

5.38 Since µ = (4)(3/10) = 1.2 and σ 2 = (4)(3/10)(7/10)(6/9) = 504/900 with σ = 0.7483, at least 3/4 of the time the number of defectives will fall in the interval

µ ± 2σ = 1.2 ± (2)(0.7483), or from − 0.297 to 2.697,

and at least 8/9 of the time the number of defectives will fall in the interval

µ ± 3σ = 1.2 ± (3)(0.7483) or from − 1.045 to 3.445.

5.39 Since µ = (13)(13/52) = 3.25 and σ 2 = (13)(1/4)(3/4)(39/51) = 1.864 with σ = 1.365, at least 75% of the time the number of hearts lay between

µ ± 2σ = 3.25 ± (2)(1.365) or from 0.52 to 5.98.

5.40 The binomial approximation of the hypergeometric with p = 1 − 4000/10000 = 0.6

gives a probability of

b(x; 15, 0.6) = 0.2131.

x=0

5.41 Using the binomial approximation of the hypergeometric with p = 0.5, the probability P 2

is 1 − b(x; 10, 0.5) = 0.9453.

x=0

5.42 Using the binomial approximation of the hypergeometric distribution with p = 30/150 =

0.2, the probability is 1 −

b(x; 10, 0.2) = 0.3222.

x=0

5.43 Using the binomial approximation of the hypergeometric distribution with 0.7, the

P 13

probability is 1 −

b(x; 18, 0.7) = 0.6077.

x=10

5.44 Using the extension of the hypergeometric distribution the probability is

5.45 (a) The extension of the hypergeometric distribution gives a probability

(b) Using the extension of the hypergeometric distribution, we have

64 Chapter 5 Some Discrete Probability Distributions

5.46 Using the extension of the hypergeometric distribution the probability is

5.50 N = 10000, n = 30 and k = 300. Using binomial approximation to the hypergeometric distribution with p = 300/10000 = 0.03, the probability of {X ≥ 1} can be determined by

1 − b(0; 30, 0.03) = 1 − (0.97) 30 = 0.599.

5.51 Using the negative binomial distribution, the required probability is

b (10; 5, 0.3) =

5.52 From the negative binomial distribution, we obtain

5.53 (a) P (X > 5) =

p(x; 5) = 1 −

p(x; 5) = 0.3840.

x=6

x=0

(b) P (X = 0) = p(0; 5) = 0.0067.

5.54 (a) Using the negative binomial distribution, we get

b ∗ (7; 3, 1/2) =

(b) From the geometric distribution, we have g(4; 1/2) = (1/2)(1/2) 3 = 1/16.

Solutions for Exercises in Chapter 5

5.55 The probability that all coins turn up the same is 1/4. Using the geometric distribution with p = 3/4 and q = 1/4, we have

3 X 3 X 63

P (X < 4) =

g(x; 3/4) =

5.56 (a) Using the geometric distribution, we have g(5; 2/3) = (2/3)(1/3) 4 = 2/243.

(b) Using the negative binomial distribution, we have

5.57 Using the geometric distribution (a) P (X = 3) = g(3; 0.7) = (0.7)(0.3) 2 = 0.0630.

(b) P (X < 4) =

g(x; 0.7) =

5.58 (a) Using the Poisson distribution with x = 5 and µ = 3, we find from Table A.2 that

p(5; 3) =

p(x; 3) −

p(x; 3) = 0.1008.

x=0

x=0

(b) P (X < 3) = P (X ≤ 2) = 0.4232. (c) P (X ≥ 2) = 1 − P (X ≤ 1) = 0.8009.

5.59 (a) P (X ≥ 4) = 1 − P (X ≤ 3) = 0.1429. (b) P (X = 0) = p(0; 2) = 0.1353.

5.60 (a) P (X < 4) = P (X ≤ 3) = 0.1512. (b) P (6 ≤ X ≤ 8) = P (X ≤ 8) − P (X ≤ 5) = 0.4015.

5.61 (a) Using the negative binomial distribution, we obtain

(b) From the geometric distribution, we have g(3; 0.8) = (0.8)(0.2) 2 = 0.032.

5.62 (a) Using the Poisson distribution with µ = 12, we find from Table A.2 that P (X < 7) = P (X ≤ 6) = 0.0458.

(b) Using the binomial distribution with p = 0.0458, we get

b(2; 3, 0.0458) =

66 Chapter 5 Some Discrete Probability Distributions

5.63 (a) Using the Poisson distribution with µ = 5, we find P (X > 5) = 1 − P (X ≤ 5) = 1 − 0.6160 = 0.3840. (b) Using the binomial distribution with p = 0.3840, we get

b(3; 4, 0.384) =

(c) Using the geometric distribution with p = 0.3840, we have

g(5; 0.384) = (0.394)(0.616) 4 = 0.0553. P 4

5.64 µ = np = (2000)(0.002) = 4, so P (X < 5) = P (X ≤ 4) ≈ p(x; 4) = 0.6288.

x=0

5.65 µ = np = (10000)(0.001) = 10, so

P (6 ≤ X ≤ 8) = P (X ≤ 8) − P (X ≤ 5) ≈

p(x; 10) −

p(x; 10) = 0.2657.

x=0

x=0

5.66 (a) µ = np = (1875)(0.004) = 7.5, so P (X < 5) = P (X ≤ 4) ≈ 0.1321. (b) P (8 ≤ X ≤ 10) = P (X ≤ 10) − P (X ≤ 7) ≈ 0.8622 − 0.5246 = 0.3376.

5.67 (a) µ = (2000)(0.002) = 4 and σ 2 = 4.

(b) For k = 2, we have µ ± 2σ = 4 ± 4 or from 0 to 8.

5.68 (a) µ = (10000)(0.001) = 10 and σ 2 = 10. √

(b) For k = 3, we have µ ± 3σ = 10 ± 3

10 or from 0.51 to 19.49.

5.69 (a) P (X ≤ 3|λt = 5) = 0.2650. (b) P (X > 1|λt = 5) = 1 − 0.0404 = 0.9596.

5.70 (a) P (X = 4|λt = 6) = 0.2851 − 0.1512 = 0.1339. (b) P (X ≥ 4|λt = 6) = 1 − 0.1512 = 0.8488.

P 74

(c) P (X ≥ 75|λt = 72) = 1 −

p(x; 74) = 0.3773.

x=0

5.71 (a) P (X > 10|λt = 14) = 1 − 0.1757 = 0.8243. (b) λt = 14.

5.72 µ = np = (1875)(0.004) = 7.5.

Solutions for Exercises in Chapter 5

5.74 µ = 1 and σ 2 = 0.99.

5.75 µ = λt = (1.5)(5) = 7.5 and P (X = 0|λt = 7.5) = e −4 = 5.53 × 10 .

5.76 (a) P (X ≤ 1|λt = 2) = 0.4060. (b) µ = λt = (2)(5) = 10 and P (X ≤ 4|λt = 10) = 0.0293.

5.77 (a) P (X > 10|λt = 5) = 1 − P (X ≤ 10|λt = 5) = 1 − 0.9863 = 0.0137. (b) µ = λt = (5)(3) = 15, so P (X > 20|λt = 15) = 1 − P (X ≤ 20|λ = 15) =

5.78 p = 0.03 with a g(x; 0.03). So, P (X = 16) = (0.03)(1 − 0.03) 15 = 0.0190 and µ= 1 0.03 − 1 = 32.33.

5.79 So, Let Y = number of shifts until it fails. Then Y follows a geometric distribution with p = 0.10. So,

P (Y ≤ 6) = g(1; 0.1) + g(2; 0.1) + · · · + g(6; 0.1)

5.80 (a) The number of people interviewed before the first refusal follows a geometric distribution with p = 0.2. So.

which is a very rare event. (b) µ = 1

5.81 n = 15 and p = 0.05.

(b) p = 0.07. So, P (X ≤ 1) =

5.82 n = 100 and p = 0.01.

(b) For p = 0.05, P (X ≤ 3) = x

x (0.05)

(1 − 0.05) 100−x = 0.2578.

x=0

68 Chapter 5 Some Discrete Probability Distributions

5.83 Using the extension of the hypergeometric distribution, the probability is

5.84 λ = 2.7 call/min.

P 4 e −2.7 (2.7) x

P 1 e −2.7 (2.7) x

(c) λt = 13.5. So,

P 10 e −13.5 (13.5) x

5.85 n = 15 and p = 0.05.

15 5 (a) P (X = 5) = 10

(b) I would not believe the claim of 5% defective.

5.86 λ = 0.2, so λt = (0.2)(5) = 1.

P 1 e −1 (1) x

(a) P (X ≤ 1) = −1

= 2e = 0.7358. Hence, P (X > 1) = 1−0.7358 = 0.2642.

x!

x=0

P 1 e −1.25 (1.25) x

(b) λ = 0.25, so λt = 1.25. P (X|le1) =

5.87 (a) 1 − P (X ≤ 1) = 1 − 100−x

x (0.05) (0.95) 100−x = 0.0371.

x=0

e −5 5 5.88 (a) 100 visits/60 minutes with λt = 5 visits/3 minutes. P (X = 0) = 0

4 1 0 5 5.89 (a) P (X ≥ 1) = 1 − P (X = 0) = 1 − 4

5.90 n = 5 and p = 0.4; P (X ≥ 3) = 1 − P (X ≤ 2) = 1 − 0.6826 = 0.3174.

5.91 (a) µ = bp = (200)(0.03) = 6. (b) σ 2 = npq = 5.82.

Solutions for Exercises in Chapter 5

e 6 (6) (c) P (X = 0) = 0

= 0.0025 (using the Poisson approximation). P (X = 0) = (0.97) 200 = 0.0023 (using the binomial distribution).

5.92 (a) p 10 q 0 = (0.99) 10 = 0.9044.

10 (b) p 10 q 12−10

5.93 n = 75 with p = 0.999. (a) X = the number of trials, and P (X = 75) = (0.999) 75 (0.001) 0 = 0.9277.

(b) Y = the number of trials before the first failure (geometric distribution), and

P (Y = 20) = (0.001)(0.999) 19 = 0.000981.

0 (c) 1 − P (no failures) = 1 − (0.001) 10 (0.999) = 0.01.

5.94 (a) 10

1 pq = (10)(0.25)(0.75) = 0.1877. (b) Let X be the number of drills until the first success. X follows a geometric distribution with p = 0.25. So, the probability of having the first 10 drills being

failure is q 10 = (0.75) 10 = 0.056. So, there is a small prospects for bankruptcy. Also, the probability that the first success appears in the 11th drill is pq 10 = 0.014

which is even smaller.

5.95 It is a negative binomial distribution. x−1 k−1 p k q x−k = 6−1 2−1 (0.25) 2 (0.75) 4 = 0.0989.

5.96 It is a negative binomial distribution. 2

5.97 n = 1000 and p = 0.01, with µ = (1000)(0.01) = 10. P (X < 7) = P (X ≤ 6) = 0.1301.

5.98 n = 500; (a) If p = 0.01,

X 14 500

P (X ≥ 15) = 1 − P (X ≤ 14) = 1 − x (0.01) (0.99) 500−x = 0.00021.

x=0

This is a very rare probability and thus the original claim that p = 0.01 is ques- tionable.

(b) P (X = 3) = 500

(c) For (a), if p = 0.01, µ = (500)(0.01) = 5. So P (X ≥ 15) = 1 − P (X ≤ 14) = 1 − 0.9998 = 0.0002.

For (b),

P (X = 3) = 0.2650 − 0.1247 = 0.1403.

5.99 N = 50 and n = 10.

70 Chapter 5 Some Discrete Probability Distributions

0 )( 10 ) (a) k = 2; P (X ≥ 1) = 1 − P (X = 0) = 1 − 50 ( = 1 − 0.6367 = 0.3633.

(b) Even though the lot contains 2 defectives, the probability of reject the lot is not very high. Perhaps more items should be sampled.

(c) µ = (10)(2/50) = 0.4.

5.100 Suppose n items need to be sampled. P (X ≥ 1) = 1 − (50−n)(49−n)

0 )( n )

n ) =1− (50)(49) ≥ 0.9. The solution is n = 34.

5.101 Define X = number of screens will detect. Then X ∼ b(x; 3, 0.8). (a) P (X = 0) = (1 − 0.8) 3 = 0.008.

(b) P (X = 1) = (3)(0.2) 2 (0.8) = 0.096.

2 (0.2) + (0.8) (c) P (X ≥ 2) = P (X = 2) + P (X = 3) = (3)(0.8) 3 = 0.896. 5.102 (a) P (X = 0) = (1 − 0.8) n ≤ 0.0001 implies that n ≥ 6.

(b) (1 − p) 3 ≤ 0.0001 implies p ≥ 0.9536.

5.103 n = 10 and p = 2 50 = 0.04.

P (X ≥ 1) = 1 − P (X = 0) ≈ 1 − (0.04) (1 − 0.04) = 1 − 0.6648 = 0.3351.

The approximation is not that good due to n N = 0.2 is too large.

5.104 (a) P = 5 = 0.1.

(b) P = 5 = 0.2.

( 2 ) 5.105 n = 200 with p = 0.00001.