Answer to Question #185307 in Statistics and Probability for Mariam

Question

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

is correct. Suppose that one of the students who takes the examination answers each of the

questions with an independent random guess. What is the probability that he answers at least ten

questions correctly?

Answer 1

The probability that a student will answer one question correctly is $p = \frac{1}{5} \Rightarrow q = \frac{4}{5}$ .

Using Bernoulli's formula, we find the probabilities that a student will correctly answer 10, 11, 12, 13, 14 and 15 questions, respectively:

$P(10) = C_{15}^{10}{p^{10}}{q^5} = \frac{{15!}}{{10!5!}} \cdot {\left( {\frac{1}{5}} \right)^{10}} \cdot {\left( {\frac{4}{5}} \right)^5} = \frac{{3075072}}{{30517578125}}$

$P(11) = C_{15}^{11}{p^{11}}{q^4} = \frac{{15!}}{{11!4!}} \cdot {\left( {\frac{1}{5}} \right)^{11}} \cdot {\left( {\frac{4}{5}} \right)^4} = \frac{{69888}}{{6103515625}}$

$P(12) = C_{15}^{12}{p^{12}}{q^3} = \frac{{15!}}{{12!3!}} \cdot {\left( {\frac{1}{5}} \right)^{12}} \cdot {\left( {\frac{4}{5}} \right)^3} = \frac{{5824}}{{6103515625}}$

$P(13) = C_{15}^{13}{p^{13}}{q^2} = \frac{{15!}}{{13!2!}} \cdot {\left( {\frac{1}{5}} \right)^{13}} \cdot {\left( {\frac{4}{5}} \right)^2} = \frac{{336}}{{6103515625}}$

$P(14) = C_{15}^{14}{p^{14}}q = \frac{{15!}}{{14!1!}} \cdot {\left( {\frac{1}{5}} \right)^{14}} \cdot \left( {\frac{4}{5}} \right) = \frac{{12}}{{6103515625}}$

$P(15) = {p^{15}} = {\left( {\frac{1}{5}} \right)^{15}} = \frac{1}{{30517578125}}$

Then the wanted probability is

$P(x \ge 10) = P(10) + P(11) + P(12) + P(13) + P(14) + P(15) = \frac{{3075072}}{{30517578125}} + \frac{{69888}}{{6103515625}} + \frac{{5824}}{{6103515625}} + \frac{{336}}{{6103515625}} + \frac{{12}}{{6103515625}} + \frac{1}{{30517578125}} = \frac{{3455373}}{{30517578125}}$

Answer: $P(x \ge 10) = \frac{{3455373}}{{30517578125}}$