# Solved (Free): About 75% of dog owners buy holiday presents for their dogs. Suppose n = 4 dog owners are randomly selected

#### ByDr. Raju Chaudhari

Mar 28, 2021

About 75% of dog owners buy holiday presents for their dogs. Suppose n = 4 dog owners are randomly selected. Find the probability that

(a) three or more buy their dog holiday presents.
(b) at most three buy their dog holiday presents
(c) Find the expected number of persons, in the sample, who buy their dog holiday presents.

#### Solution

Here $X$ denote the number of dog owners who buy their dog holiday present.

$p$ be the probability that a dog owner buy holiday present.

Given that $p=0.75$ and $n =4$. Thus$X\sim B(4, 0.75)$.

The probability mass function of $X$ is
 \begin{aligned} P(X=x) &= \binom{4}{x} (0.75)^x (1-0.75)^{4-x},\\ &\quad x=0,1,\cdots, 4. \end{aligned}

(a) The probability that three or more buy their dog holiday presents is

 \begin{aligned} P(X\geq 3) & =1-P(X\leq 2)\\ &= 1-\sum_{x=0}^{2} P(x)\\ &=1- \sum_{x=0}^{2} \binom{4}{x}(0.75)^x(1-0.75)^{4-x}\\ &=1- (0.0039+0.0469+0.2109)\\ & = 1-0.2617 \\ & = 0.7383 \\ \end{aligned}

(b) The probability that at the most three buy their dog holiday presents is
 \begin{aligned} P(X\leq 3) & =\sum_{x=0}^{3} P(x)\\ & =\sum_{x=0}^{3}\binom{4}{x}(0.75)^x(1-0.75)^{4-x}\\ &= (0.0039+0.0469+0.2109+0.4219)\\ & = 0.6836 \end{aligned}

(C) The expected number of persons, in the sample, who buy their dog holiday presents = $E(X)=n*p = 4 * 0.75 = 3$.