Question

The incidence of a certain disease is such that on the average 20% of workers suffer from it. If 10 workers are selected at random, find the probability that (i) exactly two workers suffer from the disease (ii) not more than two workers suffer from the disease.

Answer 1

i dont know i find your answer

Answer 2

Answer:

(i) P(X=2)  0.302

(ii) P(X≤2) 0.678

Step-by-step explanation:

Probability of workers suffering from a disease (p) = 20% = 0.2

Then , q= 1-p = 0.8

Number of workers being selected at random (n) = 10

We will solve it using binomial distribution,

P(X= x) = $\binom{n}{x}p^{x}q^{n-x}$

(i) P(X=2) = $\binom{10}{2}0.2^{2}0.8^{8}$

               =  0.302

(ii) P(X≤ 2) = P(X=0) + P(X=1) + P(X=2)

                =  $\binom{10}{0} 0.2^{0}0.8^{10}$ +  $\binom{10}{1} 0.2^{1}0.8^{9}$ +  $\binom{10}{2} 0.2^{2}0.8^{8}$

                = 0.107+0.269+0.302

                = 0.678