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?
• what is the probability that the student can solve at least 1 problem?

Explanation:

Answer:

• How many different exams can the instructor give?: \(C(10,5)=\frac{10!}{5!(10 - 5)!}=252\)
• What is the probability that the student can solve all 5 problems?: \(\frac{C(6,5)}{C(10,5)}=\frac{6}{252}=\frac{1}{42}\)
• What is the probability that the student can solve exactly 4 problems?: \(\frac{C(6,4)\times C(4,1)}{C(10,5)}=\frac{15\times4}{252}=\frac{60}{252}=\frac{5}{21}\)
• What is the probability that the student can solve at least 1 problem?: \(1-\frac{C(4,5)}{C(10,5)}=1 - 0 = 1\)