# Solved (Free): Richard has just been given a ten-question multiple choice test in his history class. Each question has five answers only one of which is correct

#### ByDr. Raju Chaudhari

Apr 4, 2021

Richard has just been given a ten-question multiple choice test in his history class. Each question has five answers only one of which is correct. Since Richard has not attended class recently, he does not know any of the answers. Assume that Richard guesses randomly on all ten questions. a) Find the probability that he will answer all 10 questions correctly. b) Find the probability that he will answer 5 or more questions correctly. c) Find the probability that he will answer none of the questions correctly. d) Find the probability that he will answer at least 3 questions correctly.

#### Solution

Total no. of questions $n = 10$, the probability of correct answer = 1/5 = 0.2.

Let $X$ denote the number of correct answers out of 10.

Thus $X\sim B(10, 0.2)$.

The probability mass function of $X$ is
 \begin{aligned} P(X=x) &= \binom{10}{x} (0.2)^x (1-0.2)^{10-x},\\ &\quad x=0,1,\cdots, 10. \end{aligned}

(a) The probability that he will answer all 10 questions correctly is

 \begin{aligned} P(X= 10) &= \binom{10}{10}(0.2)^{10}(1-0.2)^{0}\\ & = 0.0000001 \end{aligned}

(b) The probability that he will answer 5 or more questions correctly is

 \begin{aligned} P(X\geq 5) & =1- P(X\leq 4)\\ & = 1- \sum_{x=0}^{4} P(x)\\ & =1- \bigg(\sum_{x=0}^{4}\binom{10}{x}(0.2)^x(1-0.2)^{10-x}\bigg)\\ & =1- \bigg((0.1074)+(0.2684)+(0.302)+(0.2013)+(0.0881) \bigg)\\ & =1- 0.9672\\ & =0.0328. \end{aligned}

(c) The probability that he will answer none of the questions correctly is

 \begin{aligned} P(X= 0) &= \binom{10}{0}(0.2)^0(1-0.2)^{10}\\ & = 0.1074 \end{aligned}

(d) The probability that he will answer at least 3 questions correctly is

 \begin{aligned} P(X\geq 3) & =1- P(X\leq 2)\\ & = 1- \sum_{x=0}^{2} P(x)\\ & =1- \bigg(\sum_{x=0}^{2}\binom{10}{x}(0.2)^x(1-0.2)^{10-x}\bigg)\\ & =1- \bigg((0.1074)+(0.2684)+(0.302) \bigg)\\ & =1- 0.6778\\ & =0.3222. \end{aligned}