Answer to Question #280136 in Statistics and Probability for Aye

Question

Fourty-four percent of a corporation’s blue-collar employees

were in favor of a modified health care plan,

and 24% of its blue-collar employees favored a proposal

to change the work schedule. Thirty percent of

those favoring the health care plan modification, favored

the work schedule change.

a. What is the probability that a randomly selected

blue-collar employee is in favor of both the modified

health care plan and the changed work schedule?

b. What is the probability that a randomly chosen

blue-collar employee is in favor of at least one of

the two changes?

c. What is the probability that a blue-collar employee

favoring the work schedule change also favors the

modified health care plan?

Answer 1

Let $A$ be the event that employees are in favor of a modified heath plan and $B$ be the event that employees favored a proposal to change the work schedule.

Given $P(A)=0.44, P(B)=0.24, P(B|A)=0.30$

a)

$$P(B|A)=\dfrac{P(A\cap B)}{P(A )}$$

$$=>P(A\cap B)=P(B|A)P(A)$$

$$=0.3(0.44)=0.132$$

$$P(A\cap B)=0.132$$

b)

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

$$=0.44+0.24-0.132=0.548$$

$$P(A\cup B)=0.548$$

c)

$$P(A|B)=\dfrac{P(A\cap B)}{P(B )}=\dfrac{0.132}{0.24}=0.55$$

$$P(A|B)=0.55$$