Question

The probability that a randomly selected person has high blood pressure (the event H) is P(H) = 0.3 and the probability that a randomly selected person is a runner (the event R) is P(R) = 0.4. The probability that a randomly selected person has high blood pressure and is a runner is 0.2. Find the probability that a randomly selected person is a runner, given that he has high blood pressure.

Answer 1

Answer:

The probability of choosing a person with BP  who is also a runner is 2/3.

Step-by-step explanation:

According to the given data:

P( Selecting a  person with high blood pressure )  =  0.3

or, P(H)  = 0.3

P( Selecting a  person who is a Runner )  =  0.4

or, P(R)  = 0.4

Now, P( Randomly Selecting a person who has high BP and is a runner)  = 0.2

⇒ P(H∩ R)  = 0.2

Now, we need to find the P(randomly selected person is a runner and has already has high BP)

or we need to find: P(R/H)

now, by BAYES THEOREM:

$P(R/H) = \frac{P(R\cap H)}{P(H)}$

$\implies P(R/H) = \frac{0.2}{0.3 } = \frac{2}{3}$

Hence, the probability of choosing a person with BP  who is also a runner is 2/3.