Question

Philip ran out of time while taking a multiple-choice test and plans to guess on the last 444 questions. Each question has 555 possible choices, one of which is correct. Let X=X=X, equals the number of answers Philip correctly guesses in the last 444 questions. Assume that the results of his guesses are independent.
What is the probability that he answers exactly 1 question correctly in the last 4 questions?

Answer 1

Answer:

There is a 40.96% probability that he answers exactly 1 question correctly in the last 4 questions.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either it is correct, or it is not. This means that 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 combinatios 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:

There are four questions, so n = 4.

Each question has 5 options, one of which is correct. So $p = \frac{1}{5} = 0.2$

What is the probability that he answers exactly 1 question correctly in the last 4 questions?

This is $P(X = 1)$

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

$P(X = 1) = C_{4,1}*(0.2)^{1}*(0.8)^{3} = 0.4096$

There is a 40.96% probability that he answers exactly 1 question correctly in the last 4 questions.