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Chapter 4: Problem 38

A student prepares for an exam by studying a list of 10 problems. She can solve 6 of them. For the exam, the instructor selects 5 questions at random from the list of \(10 .\) What is the probability that the student can solve all 5 problems on the exam?

Short Answer

Expert verified
Answer: The probability that the student can solve all 5 problems on the exam is \(\frac{1}{42}\).

Step by step solution

01

Finding the number of favorable combinations

We use the formula for combinations to find the number of ways the instructor can select 5 problems from the 6 problems the student can solve. We plug in n=6 and k=5 into the formula: $$ C(6, 5) = \frac{6!}{5!(6-5)!} = \frac{6!}{5!1!} = \frac{720}{120\times1} = 6 $$ There are 6 favorable combinations. Step 2: Calculate the total number of ways the instructor can select 5 problems out of the 10.
02

Finding the total number of combinations

We use the formula for combinations to find the total number of ways the instructor can select 5 problems out of the 10. We plug in n=10 and k=5 into the formula: $$ C(10, 5) = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!} = \frac{3628800}{120\times120} = 252 $$ There are 252 total combinations. Step 3: Calculate the probability that the student can solve all 5 problems on the exam.
03

Calculate the probability

We divide the number of favorable combinations (step 1) by the total number of combinations (step 2) to find the probability: $$ P(\text{can solve all 5 problems}) = \frac{\text{favorable combinations}}{\text{total combinations}} = \frac{6}{252} = \frac{1}{42} $$ The probability that the student can solve all 5 problems on the exam is \(\frac{1}{42}\).

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