Probability question?.

Question

Here is an estimate of the number of credit-card holders. 
Age group       % of population     % of group having credit cards
______________________________________________________________________

Under 35         49.2               64.1
35-64            38.1               77.7
65 or over       12.7               68.6

If a person is selected at random, find the probability of the following.

(a) The person is under 35 and has a credit card.

I know that this is .315

(b) The person is 35–64 and has a credit card.

I know that this is .296

(c) The person is 65 or over and has a credit card.

I know that this is .087

(d) A person is selected at random from credit-card holders. Find the probability the person is in the 35–64 age group.

How can I figure this out?

Answer 1

Well, credit card holders comprise $.315372+.296037+.087122=.698531$ of the population, while credit card holders in the $35$-$64$ age group comprise $.296037$ of the population. What portion of the credit card holders are in the $35$-$64$ age group, then? (If it helps, suppose that the total population is $1,\!000,\!000$. How many credit card holders are there in that case? How many of them are in the desired age group?)

Answer 2

This solution uses Bayes' Theorem. We will also use $P(U\text{ }|\text{ }V)$ to denote the probability that the event $U$ occurs given that the event $V$ occurs.

Let $A$ be the event that a given person is under 35.

Let $B$ be the event that a given person is over 65.

Let $X$ be the event that a given person is in the 35-64 age group. Then $\neg X$ is the event that a given person is not in this age group, which is equivalent to saying that the person is either under 35 or over 65.

Thus, $P(\neg X) = P(A) + P(B)$.

Let $Y$ be the event that a given person has a credit card.

We are given the following:

$$P(A) = 49.2\%$$ $$P(B) = 12.7\%$$ $$P(X) = 38.1\%$$ $$P(Y\text{ }|\text{ }A) = 64.1\%$$ $$P(Y\text{ }|\text{ }B) = 68.6\%$$ $$P(Y\text{ }|\text{ }X) = 77.7\%$$

We wish to find the probability that a person is in the 35-65 age group, given that he/she has a credit card. We want to find $P(X\text{ }|\text{ }Y)$.

First, we find $P(Y)$. We know $$P(Y) = P(Y\text{ }|\text{ }A) \cdot P(A) + P(Y\text{ }|\text{ }B) \cdot P(B) + P(Y\text{ }|\text{ }X) \cdot P(X)$$

We can just plug in the known values to find $$P(Y) = (0.641)(0.492)+(0.686)(0.127)+(0.777)(0.381) = 0.698531$$

(Note that this is the sum of your answers to parts [a], [b], and [c])

By Bayes' Theorem, $$P(X\text{ }|\text{ }Y) = \frac{P(Y\text{ }|\text{ }X) \cdot P(X)}{P(Y)}$$

We plug in values again, to get $$P(X\text{ }|\text{ }Y) = \frac{(0.777)(0.381)}{0.698531} = \frac{296037}{698531} \approx 42.4\%$$

I hope that helps.