Question

a student prepares for an exam by studying a list of 10 different problems. she can solve 6 of the problems. for the exam, the instructor selects 5 questions at random from the list of 10.

• how many different exams can the instructor give?
• what is the probability that the student can solve all 5 problems?
• what is the probability that the student can solve exactly 4 problems?
• what is the probability that the student can solve at least 1 problem?

Explanation:

Step1: Calculate number of different exams

The number of ways to choose 5 questions out of 10 is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 10$ and $r=5$.
$C(10,5)=\frac{10!}{5!(10 - 5)!}=\frac{10\times9\times8\times7\times6}{5\times4\times3\times2\times1}=252$

Step2: Probability of solving all 5 problems

The number of ways to choose 5 problems out of the 6 she can solve is $C(6,5)=\frac{6!}{5!(6 - 5)!}=6$. The probability $P(X = 5)=\frac{C(6,5)}{C(10,5)}=\frac{6}{252}=\frac{1}{42}$

Step3: Probability of solving exactly 4 problems

The number of ways to choose 4 problems out of 6 she can solve and 1 problem out of 4 she can't solve is $C(6,4)\times C(4,1)$.
$C(6,4)=\frac{6!}{4!(6 - 4)!}=\frac{6\times5}{2\times1}=15$, $C(4,1)=\frac{4!}{1!(4 - 1)!}=4$. So $C(6,4)\times C(4,1)=15\times4 = 60$. The probability $P(X = 4)=\frac{C(6,4)\times C(4,1)}{C(10,5)}=\frac{60}{252}=\frac{5}{21}$

Step4: Probability of solving at least 1 problem

First, find the probability of solving 0 problems. The number of ways to choose 5 problems out of 4 she can't solve is $C(4,5)=0$ (since $r>n$). The probability of solving 0 problems is $P(X = 0)=0$. So the probability of solving at least 1 problem is $P(X\geq1)=1 - P(X = 0)=1$

Answer:

• The number of different exams: 252
• Probability of solving all 5 problems: $\frac{1}{42}$
• Probability of solving exactly 4 problems: $\frac{5}{21}$
• Probability of solving at least 1 problem: 1