# Solved (Free): A drug is prepared and found that when a certain quantity is inoculated into mice

#### ByDr. Raju Chaudhari

Mar 30, 2021

A drug is prepared and found that when a certain quantity is inoculated into mice, 25% of the mice become infected. If 3 mice are inoculated independently, what are the probabilities that:

a. No mice become infected;

b. One mouse becomes infected;

c. Two mice become infected;

d. All three mice become infected?

#### Solution

Let $p$ be the probability that the mouse become infected. Given that $p=0.25$. And $n=3$ mice are inoculated independently . Thus $X$ denote the number of infected mice.

Here $X\sim B(3, 0.25)$.

The probability mass function of $X$ is

 \begin{aligned} P(X=x) &= \binom{3}{x} (0.25)^x (1-0.25)^{3-x},\\ &\quad x=0,1,\cdots, 3. \end{aligned}

a. The probability that no mice become infected is

 \begin{aligned} P(X= 0) &= \binom{3}{0}(0.25)^0(1-0.25)^{3}\\ & = 0.4219 \end{aligned}

b. The probability that 1 mouse become infected is

 \begin{aligned} P(X= 1) &= \binom{3}{1}(0.25)^1(1-0.25)^{2}\\ & = 0.4219 \end{aligned}

c. The probability that 2 mice become infected is

 \begin{aligned} P(X= 2) &= \binom{3}{2}(0.25)^2(1-0.25)^{1}\\ & = 0.1406 \end{aligned}

d. The probability that all 3 mice become infected is

 \begin{aligned} P(X= 3) &= \binom{3}{3}(0.25)^3(1-0.25)^{0}\\ & = 0.0156 \end{aligned}