Question

A class receives a list of 20 study problems, from which 10 will be part of an upcoming exam. A student knows how to solve 15 of the problems. Find the probability that the student will be able to answer (a) all 10 questions on the exam, (b) exactly eight questions on the exam, and (c) at least nine questions on the exam.

Answer 1

Answer:

(a) 0.01625

(b) 0.3483

(c)  0.15170

Step-by-step explanation:

Given that there are total of 20 study problems from which 10 will come in the exam and A student knows how to solve 15 of the problems from those 20 total problems.

(a) To calculate the probability that the student will be able to answer all 10 questions on the exam, we know that:

students will be able answer all 10 questions in the exam only when all these 10 questions will be from the 15 problem which he know how to solve.

So the chances that he knows all 10 questions in the exam = $^{15}C_1_0$

And total ways in which he answer 10 question from the 20 study problems =$^{20}C_1_0$

Therefore, the Probability that the student will be able to answer all 10 questions on the exam = $\frac{^{15}C_1_0}{^{20}C_1_0}$ = $\frac{15!}{5!\times 10!} \times \frac{10!\times 10!}{20!}$ {Because $^{n}C_r$ = $\frac{n!}{r!\times (n-r)!}$ }

                                                  = 0.01625

(b) Probability that the student will be able to answer exactly eight questions on the exam = In the numerator there will be No. of ways that he answer exactly eight questions from the 15 problems he knows and the remaining 2 questions he solve from the 5 questions whose answer he doesn't know and in the denominator there will Total number of ways in which he answer 10 question from the 20 study problems.

So Required Probability = $\frac{^{15}C_8 \times ^{5}C_2 }{^{20}C_1_0 }$  = $\frac{15!}{8!\times 7!}\times \frac{5!}{2!\times 3!}\times \frac{10!\times 10!}{20!}$ = 0.3483

(c) To calculate the probability that student will be able to answer at least nine questions on the exam is given by that [He will be able to nine questions on the exam + He will be able to answer all  10 questions on the exam]

So no. of ways that he will be able to nine questions on the exam = He answer 9 questions from those 15 problems which he know and remaining one question from the other 5 questions we he don't know = $^{15}C_9\times ^{5}C_1$

And no. of ways that he will be able to answer all  10 questions on the exam

= $^{15}C_1_0$  

So, the required probability = $\frac{(^{15}C_9\times ^{5}C_1)+^{15}C_1_0}{^{20}C_1_0}$ = 0.15170