Question

1. It is known that a probability of a male child birth in a family is known to be 0.6. Find
the probability that out of 6 births recorded in the family
a) There will be exactly 3 male children.
b) There will be at least two male children
c) There will be no female child.

Answer 1

Let $X=$ the number of male children: $X\sim Bin(n, p)$

$$P(X=x)=\dbinom{n}{k}p^x(1-p)^{n-x}$$

Given $p=0.6, n=6.$

a)

$$P(X=3)=\dbinom{6}{3}0.6^3(1-0.6)^{6-3}=0.27648$$

b)

$$P(X\geq2)=1-P(X=0)-P(X=1)=$$ $$=1-\dbinom{6}{0}0.6^0(1-0.6)^{6-0}-\dbinom{6}{1}0.6^1(1-0.6)^{6-1}=$$ $$=0.95904$$

c)

$$P(X=6)=\dbinom{6}{6}0.6^6(1-0.6)^{6-6}=0.046656$$