COMP2101 / CS20S – Discrete Mathematics

Probability Distributions

1. An automotive safety engineer claims that 1 in 10 automobile accidents is due to driver

fatigue. What is the probability that at least 3 of 5 automobile accidents are due to driver

fatigue?

Solution

Binomial

b(≥3;5,0.1) = b(3;5,0.1) + b(4;5,0.1) + b(5;5,0.1)

= 5C3 x 0.13 x 0.9

5-3 + 5C4 x 0.1

4 x 0.9

5-4 + 5C5 x 0.1

5 x 0.9

5-5

= 0.00856

2. In a certain city, incompatibility is given as the legal reason in 70 percent of all divorce

cases. Find the probability that five of the next six divorce cases filed in this city will

claim incompatibility as the reason.

Solution

Binomial

b(5;6,0.7) = 6C5 x 0.75 x 0.3

6-5

= 0.3025

3. Suppose that the probability is 0.63 that a car stolen in a certain Western city will be

recovered. Find the probability that at least 8 out of 10 cars stolen in this city will be

recovered.

Solution

Binomial

b(≥8;10,0.63) = b(8;10,0.63) + b(9;10,0.63) + b(10;10,0.63)

= 10C8 x 0.638 x 0.37

10-8+10C9 x 0.63

9 x 0.37

1+10C10 x 0.63

10 x 0.37

0

= 0.2205

n 0,1,2..,for x ) -(1 C )n,b(x; x-nxn

x == θθθ

n 0,1,2..,for x ) -(1 C )n,b(x; x-nxn

x == θθθ

n 0,1,2..,for x ) -(1 C )n,b(x; x-nxn

x == θθθ

4. If the probability is 0.75 that a person will believe a rumour about the transgression of a

certain politician, find the probabilities that

a) the eighth person to hear the rumour will be the fifth to believe it;

b) the fifteenth person to hear the rumour will be the tenth to believe it.

Solution

Negative Binomial

a) b*(8;5,0.75) = 8-1C5-1 x 0.755 x 0.25

8-5

= 0.1298

b) b*(15;10,0.75) = 15-1C10-1 x 0.7510

x 0.2515-10

= 0.1101

5. An expert sharpshooter misses a target 5 percent of the time. Find the probability that she

will miss the target for the second time on the fifteenth shot.

Solution

Negative Binomial

b*(15;2,0.05) = 15-1C2-1 x 0.052 x 0.95

15-2

= 0.0180

6. In a given city 4 percent of all licensed drivers will be involved in at least one car

accident in any given year. Find the probability that among 150 licensed drivers

randomly chosen in the city

a) only five will be involved in at least one accident in any given year

b) at most three will be involved in at least one accident in any given year

Solution

Poisson or Binomial

(a) Poisson

As λ = nθ = 150 x 0.04 = 6

p(5;6) = (65 x e

-6) / 5!

= 0.161

2k 1,k k, for x ) -(1 C )k,(x;*b k-xk1-x

1-k ++== θθθ

,...2,1,0

!);(

=

=

−

x

x

exp

x λλ

λ

2k 1,k k, for x ) -(1 C )k,(x;*b k-xk1-x

1-k ++== θθθ

Binomial

b(5;150,0.04) = 150C5 x 0.045 x 0.96

145

= 0.163

(b) Only Poisson shown (either may be used but Poisson is more easily used)

p(≤ 3;6) = p(0;6) + p(1;6) + p(2;6) + p(3;6)

= (60 x e

-6) / 0! + (6

1 x e

-6) / 1! + (6

2 x e

-6) / 2! + (6

3 x e

-6) / 3!

= 0.151

n 0,1,2..,for x ) -(1 C )n,b(x; x-nxn

x == θθθ

,...2,1,0

!);(

=

=

−

x

x

exp

x λλ

λ