Question

A small town has 7000 adult males and 5000 adult females. A sociologist conducted a survey and found that 60% of the males and 50% of the females drink heavily. An adult is selected at random from the town. (Enter your probabilities as fractions.)

(a) What is the probability the person is a male?


(b) What is the probability the person drinks heavily?


(c) What is the probability the person is a male or drinks heavily?


(d) What is the probability the person is a male, if it is known that the person drinks heavily?

Two balls are drawn from a bag containing 4 white balls and 3 red balls. If the first ball is replaced before the second is drawn, what is the probability that the following will occur? (Enter your probabilities as fractions.)

(a) both balls are red


(b) both balls are white


(c) the first ball is red and the second is white


(d) one of the balls is black

Answer 1

We are given here that:
P( males ) = 7/12
P( females) = 5/12

P( drink | males ) = 0.6 and P( drink | females ) = 0.5

a) Probability that the person is a male is computed as:
= Number of males / Total

= 7000/12000

= 7/12

Therefore 7/12 is the required probability here.

b) Using law of total probability, we get here:
P( drink) = P(drink | males)P(males) + P(drink | females)P(females)

P(drink ) = 0.6*(7/12) + 0.5*(5/12) = 0.5583

Therefore 0.5583 is the required probability here.

c) P(male or drinks heavily )

= P(male) + P(drink) - P(male and drink)

= (7/12) + 0.5583 - 0.6*(7/12)

= 0.7917

Therefore 0.7917 is the required probability here.

d) Now the probability the person is a male, if it is known that the person drinks heavily is computed using Bayes theorem as:

P( male | drink ) = P(drink | males)P(males) / P(drink )

= 0.6*(7/12) / 0.5583

= 0.6269

Therefore 0.6269 is the required probability here.