Question

A student prepares for an exam by studying a list of 10 problems. She can solve 8 of them. For the exam, the instructor selects 7 questions at random from the list of 10. What is the probability that the student can solve all 7 problems on the exam? (Enter your probability as a fraction.)

Answer 1

Answer:

There is a $\frac{2097}{10000}$ probability that she can solve all 7 problems.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he gets it correct, or he gets it wrong. So we solve this problem using the binomial probability distribution.

Binomial probability distribution

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

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

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 $\pi$ is the probability of X happening.

For this problem, we have that:

In our sample there are 7 questions, so $n = 7$.

There are 10 problems, and she can solve 8 of them. So $\pi = \frac{8}{10} = 0.5$

What is the probability that the student can solve all 7 problems on the exam?

This is $P(X = 7)$.

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

$P(X = 7) = C_{7,7}.(0.80)^{7}.(0.2)^{0} = 0.2097 = \frac{2097}{10000}$

There is a $\frac{2097}{10000}$ probability that she can solve all 7 problems.