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

Step-by-step 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 = $\dfrac{1}{3}$

Sample size : n=6

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

$P(3)=^6C_3(\dfrac{1}{3})^3(1-\dfrac{1}{3})^{6-3}\\\\=\dfrac{6!}{3!3!}(\dfrac{1}{3})^3(\dfrac{2}{3})^{3}\\\\=0.219478737997\approx0.2195$

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