Several million lottery tickets are sold, and 60% of the Tickets are held by women. Five winning tickets will be drawn at random.

a. What is the probability that three or fewer of the winners will be women?

b. What is the probability that three of the winners will be of one gender and two of the winners will be of the other gender?

Answer :

Step 1 of 3:

Given, 60% of the tickets are held by women in several million lottery tickets and 5 winning tickets are drawn at random.

Let X follows binomial distribution with probability mass function is

P(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.

Where,

n = sample size

= 5

x = random variable

p = probability of success

= 60%

= 0.60

q = 1 - p (probability of failure)

Mean of the binomial distribution is

= np

Variance of the binomial distribution is

= npq

Step 2 of 3:

The claim is to find the probability that three or fewer of the winners will be women.We have to find P(X 3) = P(x = 0) + P( x = 1) + P(x = 2) + P(x = 3)

P(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.

Where, n = 5, p = 0.6 and x = 0, 1, 2 and 3.

P(X 3) = (1 - 0.60+ (1 - 0.60+

(1 - 0.60+ (1 - 0.60

= (1) (1) (0.01024) + (5) (0.60) (0.0256) +

(10) (0.36) (0.064) + (10) (0.216) (0.16)

= 0.01024 + 0.0768 + 0.2304 + 0.3456

= 0.66304

Hence, the probability that three or fewer of the winners will be women is 0.66304