Question

Of the members of the Spring Lake Bowling Lanes, 57% have a lifetime membership

Answer 1

Answer:

The probability that a randomly selected member is a lifetime member,  given that he or she bowls regularly is 0.814, option b

Step-by-step explanation:

Let "M" denotes the event that a club member has a lifetime membership and "B" be denotes the event that the club member bowls.

57% of the members have lifetime membership and bowl regularly. This means, the probability that a randomly selected member has a lifetime membership and bowls regularly = $P(M \cap B)=57\%=0.57$

70% of the club members bowl. This means, the probability that a randomly chosen member bowls = P(B) = 70% = 0.70

We have to find the probability that a randomly selected member is a lifetime member,  given that he or she bowls regularly. This is conditional probability which can be expressed mathematically as P(M | B)

The formula to calculate the conditional probability is:

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

So, the formula for  our case will be:

$P(M|B)=\frac{P(M \cap B)}{P(B)}$

Using the values in this formula we get:

$P(M|B)=\frac{0.57}{0.70}=0.814$

This means, the probability that a randomly selected member is a lifetime member,  given that he or she bowls regularly is 0.814