Respuesta :
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
= [tex]^{7}C_5[/tex]
and total ways in which he answer 5 problems from the set of 10 problems
= [tex]^{10}C_5[/tex]
So, the Probability that he or she will answer correctly to all 5 problems
= [tex]\frac{^{7}C_5}{^{10}C_5}[/tex] = [tex]\frac{7!}{5!\times 2!}\times \frac{5!\times 5!}{10!}[/tex] = 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 = [tex]^{7}C_4 \times ^{3}C_1[/tex]
And the no. of ways that he or she will answer correctly all 5 questions in the final exam = [tex]^{7}C_5[/tex]
Therefore, the required probability = [tex]\frac{(^{7}C_4 \times ^{3}C_1)+^{7}C_5}{^{10}C_5}[/tex] =[tex]((\frac{7!}{4!\times 3!}\times \frac{3!}{1!\times 2!})+\frac{7!}{5!\times 2!})\times \frac{5!\times 5!}{10!}[/tex] = 0.5 .
