Question

In a company an average of 3 out of every 5 staffs take MC on daily basis. A random sample of 10 staffs is selected. Find the probability that Exactly 6 take MC in a particular week.

Answer 1

Let $X=$ the number of staffs taking MC: $X\sim Bin(n, p).$

Given $n=10, p=\dfrac{3}{5}=0.6$

$$P(X=6)=\dbinom{10}{6}(0.6)^6(1-0.6)^{10-6}$$

$$=210(0.6)^6(0.4)^{4}=0.250822656$$

The probability that Exactly 6 take MC in a particular week is 0.2508.