Question

In a certain town, 30% of the people own dogs. (So the probability of a person being a dog-owner is 30%).

Answer 1

From the question,

We are given that the probability of a person being a dog-owner is 30%

Therefore,

The probability of a person not being a dog-owner is 70%

Hence

probability of success, p= 30%

Probability of failure, q = 70%

By binomial probability formula we have

$p(x)=^nC_xp^{x^{}}q^{n-x}$

Part A

We are to find the probability of exactly 4 dog-owners if 10 dog owners are randomly selected

This implies

x = 4, n = 10

Applying the formula

$\begin{gathered} p(4)=^{10}C_4(0.3)^4(0.7)^{10-4} \\ P(4)=210\times\frac{81}{10000}\times0.1176 \\ p(4)=0.2 \end{gathered}$

Therefore, the probability that exactly 4 own dogs is 0.2

Part B

We are to find the probability that at least 4 of them own dogs

at least 4 means 4 and above

Therefore

The probability of at least 4 will be

$\begin{gathered} p(x\ge4)=1-p(x<4) \\ p(x\ge4)=1-\lbrack(p(x=0)+p(x=1)+p(x=2)+p(x=3)\rbrack \end{gathered}$

By substituting values we have

$p(x\ge4)=1-\lbrack^{10}C_0p^0q^{10}+^{10}C_1p^1q^9+^{10}C_2p^2q^8+^{10}C_3p^3q^7\rbrack$

Hence, we have

$\begin{gathered} p(x\ge4)=1-\lbrack1\times1\times0.028^{}+10\times0.3\times0.04+45\times0.09\times0.057+120\times0.027\times0.082\rbrack \\ p(x\ge4)=1-\lbrack0.028+0.12+0.231+0.266\rbrack \\ p(x\ge4)=1-0.645 \\ p(x\ge4)=0.355 \end{gathered}$

Therefore, the probability that at least 4 of them own dogs is 0.355