Question

A voter survey has determined that 20% of the residents of Flaugamuck County own an iPod. Furthermore, 75% of these iPod owners are under 30 years of age, whereas 25% are 30 or older. Only 35% of the people who don't own iPods are under 30 years of age.

What percentage of the residents who are 30 or older own iPods? Or, equivalently, what is the conditional probability that a resident known to be 30 or more years of age will own in iPod? (Give your answer as a percentage correct to two decimal places.)

Answer 1

Answer:

8.77% probability that a resident known to be 30 or more years of age will own in iPod

Step-by-step explanation:

We use the conditional probability formula to solve this question.

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

In which

P(B|A) is the probability of B happening when A happens

$P(A \cap B)$ is the probability of both events happening

P(A) is the probability of event A happening.

In this problem:

Event A: Over 30 years of age.

Event B: Owning an iPod.

Being over 30 years of age:

25% of those 20% who own an iPod.

100-35 = 65% of those 80% who do not own an iPod. So

$P(A) = 0.25*0.2 + 0.65*0.8 = 0.57$

Being over 30 years of age and owning an iPod:

25% of those 20% who own an iPod. So

$P(A \cap B) = 0.2*0.25 = 0.05$

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.05}{0.57} = 0.0877$

8.77% probability that a resident known to be 30 or more years of age will own in iPod