# Solved (Free): A total of 16 mice are sent down a maze, one by one. From previous experience, it is believed that

#### ByDr. Raju Chaudhari

Apr 4, 2021

A total of 16 mice are sent down a maze, one by one. From previous experience, it is believed that the probability a mouse turns right is .38. Suppose their turning pattern follows a binomial distribution.

1. What is the probability that exactly 8 of the 16 mice turn right?
2. What is the probability that 8 or fewer of the 16 mice turn right?
3. What is the probability that 8 or more turn right?
4. What is the probability that more than 3, but fewer than 10 turn right?
5. What is the probability that exactly 10 turn left?

#### Solution

Let $X$ denote the number of mice who turn right out of 16. Then $X$ take the values $X=0,1,2\cdots, 16$. The probability that a mouse trun right is $0.38$.

Given that $X\sim B(16, 0.38)$.

The probability mass function of $X$ is

 \begin{aligned} P(X=x) &= \binom{16}{x} (0.38)^x (1-0.38)^{16-x},\\ &\quad x=0,1,\cdots, 16. \end{aligned}

1. The probability that $X$ is exactly equal to $8$ is

 \begin{aligned} P(X= 8) & =\binom{16}{x}(0.38)^{8}(1-0.38)^{16-8}\\ & = 0.1222 \end{aligned}

1. The probability that $X$ is less than or equal to $8$ is

 \begin{aligned} P(X\leq 8) & =\sum_{x=0}^{8} P(x)\\ & =\sum_{x=0}^{8}\binom{16}{x}(0.38)^x(1-0.38)^{16-x}\\ & = (0.0005)+(0.0047)+(0.0215)+(0.0615)+(0.1224) \\ &\quad + (0.1801)+(0.2024)+(0.1772)+(0.1222)\\ & = 0.8924 \end{aligned}

1. The probability that $X$ is greater than or equal to $8$ is

 \begin{aligned} P(X\geq 8) & =1- P(X\leq 7)\\ & = 1-\sum_{x=0}^{7} P(x)\\ & =1-\sum_{x=0}^{7}\binom{16}{x}(0.38)^x(1-0.38)^{16-x}\\ & = 1-\bigg(0.0005+0.0047+0.0215+0.0615\\ &\quad + 0.1224+0.1801+0.2024+0.1772\bigg) \\ & = 0.2298 \end{aligned}

1. The probability that $X$ is greater than $3$ but less than $10$ is

 \begin{aligned} P(3 < X < 10) & = P(4\leq X\leq 9)\\ & = \sum_{x=4}^{9} P(x)\\ & =\sum_{x=4}^{9}\binom{16}{x}(0.38)^x(1-0.38)^{16-x}\\ & = \bigg(0.1224+0.1801+0.2024\\ &\quad + 0.1772+0.1222+0.0666\bigg) \\ & = 0.8708 \end{aligned}

1. The probability that exactly 10 turns left (means remaining 6 turns right) is

 \begin{aligned} P(X= 6) & =\binom{16}{x}(0.38)^{6}(1-0.38)^{16-6}\\ & = 0.2024 \end{aligned}