Question

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

Answer 1

Answer:

(a)  0.0833

(b) 0.5

Step-by-step explanation:

We are given that an instructor gives her class a set of 10 problems from which random selection of 5 questions will come in the final exam and a student knows how to solve 7 of the problems from those 10 total problems.

(a) To calculate the probability that the student will be able to answer all 5 problems in the final exam, we consider that:

Student will be able answer all 5 problems in the final exam only when these 5 problems came from those 7 questions which he knows how to solve.

So the ways in which he answer all 5 problems correctly in the final exam      

                      =  $^{7}C_5$

and total ways in which he answer 5 problems from the set of 10 problems

                      = $^{10}C_5$

So, the Probability that he or she will answer correctly to all 5 problems

                           = $\frac{^{7}C_5}{^{10}C_5}$  = $\frac{7!}{5!\times 2!}\times \frac{5!\times 5!}{10!}$ = 0.0833

                                               

(b) Now to calculate the probability that he or she will answer correctly at least 4 of the problems in the final exam is given by that [He or she will be able to answer correctly 4 problems in the exam + He or she will be able to answer correctly all 5 problems in the exam]

So the no. of ways that he or she will be able to answer correctly 4 problems in the exam = He or she answer correctly 4 questions from those 7 problems which he knows how to solve and the remaining one question from the other 3 questions we he don't know to solve = $^{7}C_4 \times ^{3}C_1$

And the no. of ways that he or she will answer correctly all 5 questions in the final exam = $^{7}C_5$

Therefore, the required probability =  $\frac{(^{7}C_4 \times ^{3}C_1)+^{7}C_5}{^{10}C_5}$ =$((\frac{7!}{4!\times 3!}\times \frac{3!}{1!\times 2!})+\frac{7!}{5!\times 2!})\times \frac{5!\times 5!}{10!}$ = 0.5 .