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?

Answer 1

Answer:

a) 0.02% probability that he will answer all questions correctly.

b) 82.20% probability that he will answer at least one questions correctly

c) 16.94% probability that he will answer at least half questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either Richard answer it correctly, or he answers it wrong. The probability of answering a question correctly is independent from other questions. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

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

This means that $n = 6$

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.

This means that $p = \frac{1}{4} = 0.25$

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

This is $P(X = 6)$.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{6,6}.(0.25)^{6}.(0.75)^{0} = 0.0002$

0.02% probability that he will answer all questions correctly.

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

Either he does not answer any of the questions correctly, or he does answer at least one correctly. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$

So

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.25)^{0}.(0.75)^{6} = 0.1780$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1780 = 0.8220$

82.20% probability that he will answer at least one questions correctly

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

$P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{6,3}.(0.25)^{3}.(0.75)^{3} = 0.1318$

$P(X = 4) = C_{6,4}.(0.25)^{4}.(0.75)^{2} = 0.0330$

$P(X = 5) = C_{6,5}.(0.25)^{5}.(0.75)^{1} = 0.0044$

$P(X = 6) = C_{6,6}.(0.25)^{6}.(0.75)^{0} = 0.0002$

$P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.1318 + 0.0330 + 0.0044 + 0.0002 = 0.1694$

16.94% probability that he will answer at least half questions correctly