Question

A student answers a multiple choice examination with two questions that have four possible answers each. Suppose that the probability that the student knows the answer to a question is 0.80 and the probability that the student guesses is 0.20. If the student guesses, the probability of guessing the correct answer is 0.25. The questions are independent, that is, knowing the answer on one question is not influenced by the other question. (a) What is the probability that the both questions will be answered correctly

Answer 1

Answer:

72.25% probability that the both questions will be answered correctly.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the answer is correct, or it is not. 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.

In this problem we have that:

The probability that the student knows the answer to a question is 0.80 and the probability that the student guesses is 0.20. If the student guesses, the probability of guessing the correct answer is 0.25.  So the probability that he answers a question right is $p = 0.8 + 0.2*0.25 = 0.85$.

(a) What is the probability that the both questions will be answered correctly

This is P(X = 2). So

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

$P(X = 2) = C_{2,2}.(0.85)^{2}.(0.15)^{0} = 0.7225$

There is a 72.25% probability that the both questions will be answered correctly.