Twelve employees are randomly selected from an engineering department. Ten of th

Question

Twelve employees are randomly selected from an engineering department. Ten of the employees are men and 2 are women. Two employees are randomly selected from the sample with no replacement between the draws. a. Draw a tree diagram and label all branches and probabilities. b. What is the probability that there would be at least one woman? c. What is the probability that there will be no women? d. What is the probability that there will be two women? e. What is the probability that there will be exactly one man and one woman?

Explanation / Answer

Given that there are 12 employees are randomly selected from an engineering department.

Of which 10 are men and 2 are women.

There are only two catogaries men and women so :

P(men) = 1/2 and P(women) = 1/2

Let X be the number of men and Y be the number of womens.

Distribution of X and Y follows Binomial distribution.

X ~ Binomial(n=10, p = 1/2)

The pmf of X=x is,

P(X=x) = (10 C x) * 1/2x * (1 - 1/2)(10-x)  

Y ~ Binomial(n=2,p = 1/2)

The pmf of Y=y is,

P(Y=y) = (2 C y) * 1/2y * (1 - 1/2)(2-y)  

b) Here we have to find P(Y>=1)

P(Y>=1) = 1 - P(Y<1) = 1 - P(Y=0)

P(Y=0) =

The pmf of X=x is,

P(Y=0) = (10 C 0) * 1/20 * (1 - 1/2)(2-0) = 0.25

P(Y>=1) = 1 - 0.25 = 0.75

c) Here we have to find P(Y=0)

P(Y=0) = 0.25

d) Here we have to find P(Y=2)

P(Y=2) =  (2 C 2) * 1/22 * (1 - 1/2)(2-2) = 0.25

e) Here we have to find P(X=1 and Y = 1)

P(X=1) = (10 C 1) * 1/21 * (1 - 1/2)(10-1) = 0.0098

P(Y=1) = (2 C 1) * 1/21 * (1 - 1/2)(2-1) = 0.5

P(X=1 and Y = 1) = P(X=1) * P((Y=1) = 0.0098*0.5 = 0.0049

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