Question

In a hospital, there are 20 kidney dialysis machines and the chance of any one of them to be out of service during a day is 0.02. Determine the probability that exactly 3 machines will be out of service on the same day.

Answer 1

\[\text{ Let X denote the number of machines out of service during a day . }  \]
\[\text{ Then, X follows a binomial distribution with n = 20 } \]
\[\text{ Let p be the probability of any machine out of service during a day } . \]
\[ \therefore p = 0 . 02 \text{ and }  q = 0 . 98 \]
\[\text{ Hence, the distribution is given by } \]
\[P(X = r) = ^{20}{}{C}_r \left( 0 . 02 \right)^r \left( 0 . 98 \right)^{20 - r} , r = 0, 1, 2 . . . . . 20\]
\[ \therefore P(\text{ exactlly 3 machines will be out of the service on the same day } ) = P(X = 3)\]
\[ = ^{20}{}{C}_3 \left( 0 . 02 \right)^3 \left( 0 . 98 \right)^{20 - 3} \]
\[ = 1140(0 . 000008)(0 . 7093)\]
\[ = 0 . 006469\]