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: 0.2195Step-by-step explanation:Binomial distribution formula :-[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], 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 = [tex]\dfrac{1}{3}[/tex]Sample size : n=6Now, the probability that exactly three of them have blood type A is given by :-[tex]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[/tex]Therefore, the probability that exactly three of them have blood type A = 0.2195