Question

A multiple-choice examination has 15 questions, each with five answers, only one of which is correct. Suppose that one of the students who takes exam and answers each of the questions with an independent random guess. What is the probability that he answers at least 10 questions correctly?

Answer 1

Answer:

0.0111% probability that he answers at least 10 questions correctly

Step-by-step explanation:

For each question, there are only two outcomes. Either it is answered correctly, or it is not. The probability of a question being answered correctly is independent from other questions. So we use the binomial probability distribution to solve this question.

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.

A multiple-choice examination has 15 questions, each with five answers, only one of which is correct.

This means that $n = 15, p = \frac{1}{5} = 0.2$

What is the probability that he answers at least 10 questions correctly?

$P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

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

$P(X = 10) = C_{15,10}.(0.2)^{10}.(0.8)^{5} = 0.0001$

$P(X = 11) = C_{15,11}.(0.2)^{11}.(0.8)^{4} = 0.000011$

$P(X = 12) = C_{15,12}.(0.2)^{12}.(0.8)^{3} \cong 0$

$P(X = 13) = C_{15,13}.(0.2)^{13}.(0.8)^{2} \cong 0$

$P(X = 14) = C_{15,14}.(0.2)^{14}.(0.8)^{1} \cong 0$

$P(X = 15) = C_{15,15}.(0.2)^{15}.(0.8)^{0} \cong 0$

$P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.0001 + 0.000011 = 0.000111$

0.0111% probability that he answers at least 10 questions correctly