Answer: 0.27648
Step-by-step explanation:
Given : The proportion of all individuals have group A blood : p=0.040
Total individuals give blood : n= 6
Let X be the number of individuals have group A blood.
Since all individual are independent of each other.
[tex]X\sim Bin(n=6, p=0.40)[/tex]
Formula : [tex]P(X=x)=^nC_xp^x(1-p)^{n-x}[/tex] , where n= sample size , p = probability of getting success in each trial.
The probability that exactly three of the individuals have group A blood. :
[tex]P(x=3)= ^6C_3(0.40)^3(1-0.4)^3\\\\= \dfrac{6!}{3!3!}\times(0.40)^3(0.60)^3\\\\=0.27648[/tex]
The probability that exactly three of the individuals have group A blood. is 0.27648
