# Richard had just been given a 6-question multiple-choice quiz in his history class. Each question...

## Question:

Richard had just been given a 6-question multiple-choice quiz in his history class. Each question has four answer, of which only one is correct. Since Richard had not attend not attended class recently, he doesn't any of the answer. Assuring that Richard guesses on all six questions, find the indicated probabilities.

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

b. What is the probability that he will answer at least one questions correctly?

c. What is the probability that he will answer at least half questions correctly?

## Binomial distribution:

In the experiment, binomial distribution can be applied if the number of cases (trials) is of discrete nature and independent. The probability of an expected outcome in the experiment, which is known as success, remains the same as per cases.

Given information:

• The number of questions (n) is 6.
• The number of options in each questions is 4, so the probability of success (p) is {eq}\dfrac{1}{4} = 0.25 {/eq}.

Let X be the number of questions answered correctly. So, the random variable {eq}X \sim B\left( {n,p} \right) {/eq}.

The probability density function of binomial distribution is:

{eq}P\left( {X = x} \right) = {}^n{C_x}{p^x}{\left( {1 - p} \right)^{n - x}} {/eq}

(a)

The probability that he will answer all questions correctly can be calculated as,

{eq}\begin{align*} P\left( {X = 6} \right) &= {}^6{C_6}{\left( {0.25} \right)^6}{\left( {1 - 0.25} \right)^{6 - 6}}\\ &= \dfrac{{6!}}{{6!\left( {6 - 6} \right)!}}{\left( {0.25} \right)^6}{\left( {0.75} \right)^0}\\ &= 0.000244 \end{align*} {/eq}

Therefore, the probability that he will answer all questions correctly is 0.000244.

(b)

The probability that he will answer at least one question correctly can be calculated as,

{eq}\begin{align*} P\left( {X \ge 1} \right) &= 1 - P\left( {X < 1} \right)\\ &= 1 - P\left( {X = 0} \right)\\ &= 1 - {}^6{C_0}{\left( {0.25} \right)^0}{\left( {1 - 0.25} \right)^{6 - 0}}\\ & = 1 - 0.17797\\ &= 0.82203 \end{align*} {/eq}

Therefore, the probability that he will answer at least one question correctly is 0.82203.

(c)

The probability that he will answer at least half questions correctly can be calculated as,

{eq}\begin{align*} P\left( {X \ge 3} \right) &= P\left( {X = 3} \right) + P\left( {X = 4} \right) + P\left( {X = 5} \right) + P\left( {X = 6} \right)\\ & = {}^6{C_3}{\left( {0.25} \right)^3}{\left( {1 - 0.25} \right)^{6 - 3}} + {}^6{C_4}{\left( {0.25} \right)^4}{\left( {1 - 0.25} \right)^{6 - 4}} + {}^6{C_5}{\left( {0.25} \right)^5}{\left( {1 - 0.25} \right)^{6 - 5}} + {}^6{C_6}{\left( {0.25} \right)^6}{\left( {1 - 0.25} \right)^{6 - 6}}\\ & = 0.131835 + 0.032958 + 0.004394 + 0.000244\\ & = 0.169431 \end{align*} {/eq}

Therefore, the probability that he will answer at least half questions correctly is 0.169431.