# Richard has just been given a 8-question multiple-choice quiz in his history class. Each question...

## Question:

Richard has just been given a 8-question multiple-choice quiz in his history class. Each question has five answers, of which only one is correct. Since Richard has not attended class recently, he doesn't know any of the answers. Assuming that Richard guesses on all eight questions, find the indicated probabilities. (Round your answers to three decimal places.)

(a) What is the probability that he will answer all questions correctly?

(b) What is the probability that he will answer all questions incorrectly?

(c) What is the probability that he will answer at least one of the questions correctly? Compute this probability two ways. First, use the rule for mutually exclusive events and the probabilities shown in the binomial probability distribution table. Then use the fact that {eq}P(r \geq 1) = 1 - P(r = 0). {/eq}

(d) What is the probability that Richard will answer at least half the questions correctly?

## Mutually exclusive events and binomial distribution:

Mutually exclusive events are events that cannot occur at the same time. Binomial distribution events are mutually exclusive since such events have two possible outcomes, success or failure that cannot occur at the same time.

Let X be a random a random variable that denotes the number of questions Richard guesses correctly.

X follows binomial distribution with n = 8 and p = {eq}\dfrac{1}{5} = 0.2 {/eq}

Probability mass function of a binomial distribution is given by:{eq}P(X=x) =\begin{align*}{n \choose x} \times p^x \times (1-p)^{n-x}\end{align*}\space,x=0,1,2,3,...,n {/eq}, where;
{eq}n {/eq} is the number of trials,
{eq}p {/eq} is the probability of success,
{eq}x {/eq} is the number of successes and
{eq}\begin{align*}{n \choose x} \end{align*}= \dfrac{n!}{x!(n-x)!} {/eq} is the number of combinations of {eq}n {/eq} objects, taken {eq}x {/eq} at a time.

(a)

Probability that he will answer all questions correctly:

{eq}P(X=8) =\begin{align*}{8 \choose 8} \times 0.2^8 \times (1-0.2)^{8-8}\end{align*} = 0.00000256 {/eq}

(b)

Probability that he will answer all questions incorrectly:

{eq}P(X=0) =\begin{align*}{8 \choose 0} \times 0.2^0 \times (1-0.2)^{8-0}\end{align*} = 0.168 {/eq}

(c)

Probability that he will answer at least one of the questions correctly:

i

Using the rule for mutually exclusive events:

=1 - P(All questions are answered incorrectly)
{eq}=1-\bigg[(1 -0.2)\times (1 -0.2)\times (1 - 0.2)\times (1-0.2)\times (1-0.2) \times (1-0.2)\times (1-0.2)\times (1-0.2)\bigg]\\=1 - 0.16777\\= 0.832 {/eq}

(ii)

Using binomial probability distribution table.

{eq}P(X\geq 1) = 1 - P(X <1)\\=1 - P(X=0)\\=1 -0.168 \\= 0.832 {/eq}

(d)

Probability that Richard will answer at least half the questions correctly:

{eq}P(X\geq 4) = P(X=4)+P(X=5)+P(X=6)=P(X=7)+P(X=8)\\=\begin{align*}{8 \choose 4} \times 0.2^4 \times (1-0.2)^{8-4}+{8 \choose 5} \times 0.2^5 \times (1-0.2)^{8-5}+{8 \choose 6} \times 0.2^6 \times (1-0.2)^{8-6}+{8 \choose 7} \times 0.2^7 \times (1-0.2)^{8-7}+{8 \choose 8} \times 0.2^8 \times (1-0.2)^{8-8}\end{align*}\\=0.0458752+0.009175+0.00114688+0.00008192+0.00000256\\=0.056 {/eq}