Question

Based on data from a statistical abstract, only about 16% of senior citizens (65 years old or older) get the flu each year. However, about 30% of the people under 65 years old get the flu each year. In the general population, there are 14% senior citizens (65 years old or older). (Round your answers to three decimal places.)
What is the probability that a person selected at random from the general population is senior citizen who will get the flu this season?

Answer 1

Answer:

Probability that a person selected at random from the general population is senior citizen who will get the flu this season = 0.0224 .

Step-by-step explanation:

We are given that in the general population, there are 14% senior citizens (65 years old or older).

Let Probability of senior citizens (65 years old or older) in the population, P(S) = 0.14

Probability of citizens under 65 years old in the population, P(NS) = 1 - 0.14 = 0.86

Now, let event F = citizens getting flu each year

Probability of senior citizens (65 years old or older) getting flu each year, P(F/S) = 0.16

Probability of citizens under 65 years getting flu each year, P(F/NS) = 0.30

So, probability that a person selected at random from the general population is senior citizen who will get the flu this season = Probability that person is senior citizen(65 years old or older) * Probability of senior citizens (65 years old or older) getting flu each year

= P(S) * P(F/S) = 0.14 * 0.16 = 0.0224.

Answer 2

Answer: The required probability is 2.24%.

Step-by-step explanation:

Since we have given that

Percentage of people are senior citizens ( 65 years old or older ) = 14%

Percentage of people are under 65 years old = 100-14 = 86%

Probability that senior citizens get the flu each year = 16%

Probability that under 65 years old get the flu each year = 30%

So, Probability that a person selected at random from the general population is senior citizen who get the flu this season is given by

$P(S\cap F)={P(F|S)\times P(S)\\\\P(S\cap F)=0.16\times 0.14\\\\P(S\cap F)=0.0224\\\\P(S\cap F)=2.24\%$Hence, the required probability is 2.24%.