Question

It is estimated that one third of the general population has blood type A A sample of six people is selected at random. What is the probability that exactly three of them have blood type A?

Answer 1

Answer: 0.2195

Explanation:

Binomial distribution formula :-

`P(x)=^nC_xp^x(1-p)^(n-x)`, where P(x) is the probability of x successes in the n independent trials of the experiment and p is the probability of success.

Given : The probability of that the general population has blood type A =
`(1)/(3)`

Sample size : n=6

Now, the probability that exactly three of them have blood type A is given by :-

`P(3)=^6C_3((1)/(3))^3(1-(1)/(3))^(6-3)\\\\=(6!)/(3!3!)((1)/(3))^3((2)/(3))^(3)\\\\=0.219478737997\approx0.2195`

Therefore, the probability that exactly three of them have blood type A = 0.2195