Question

An instructor gives her class a set of 10 problems with the information that the final exam will consist of a random selection of 5 of them. If a student has figured out how to do 7 of the 10 problems, what is the probability that he will answer correctly (a) all 5 problems? (b) at least 4 of the problems?

Answer 1

Answer with explanation:

It is given that An instructor gives her class a set of 10 problems.

The final exam will consist of a random selection of 5 of them.

A student has figured out how to do 7 of the 10 problems.

We know that the probability of an event is defined as the ratio of  number of favorable outcomes to the total number of outcomes.

a)

The probability that he will answer all the 5 problems correctly is:

    given by:

$\dfrac{7_C_5}{{10}_C_5}$

Since he may chose 5 questions out of 7 questions that a student knows how to do and total outcomes is the selection of 5 questions from the total 10 questions.

Hence, on solving we get:

$=\dfrac{\dfrac{7!}{5!\times (7-5)!}}{\dfrac{10!}{5!\times (10-5)!}}\\\\\\\\=\dfrac{\dfrac{7!}{5!\times 2!}}{\dfrac{10!}{5!\times 5!}}$

On further solving we get:

Probability=0.083

b)

The probability that he will answer at least four correctly is given by:

                 $\dfrac{7_C_4\times 3_C_1+7_C_5}{10_C_5}$

( Since, he may answer correctly four or five questions)

( If he answer four answers correctly, this means he choose 4 questions from 7 questions and 1 from the remaining.

and if he answers five answer correctly then he choose it from the 7 questions a student know)

Hence, on solving we get:

$Probability=\dfrac{2}{9}\\\\\\Probability=0.222$

Hence, we get Probability= 0.222