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Chapter 2: Problem 37

An instructor gives her class a set of 10 problems with the information that the final exam will consist of a random selection of 5 of them. If a student has figured out how to do 7 of the problems, what is the probability that he or she will answer correctly (a) all 5 problems? (b) at least 4 of the problems?

Short Answer

Expert verified
(a) The probability of correctly answering all 5 problems is approximately \(8.33\%\). (b) The probability of correctly answering at least 4 problems is \(50\%\).

Step by step solution

01

(Step 1: Find Total Ways to Select 5 Problems)

The total ways to select 5 problems from a set of 10 problems can be calculated using combinations, denoted as \(C(n,k)\), where n is the total number of problems and k is the number of problems to be selected(counted). \[ C(n,k) = \frac{n!}{k!(n-k)!} \] Plugging the values \(n = 10\) and \(k=5\) to calculate the total ways to select 5 problems: \[ C(10,5) = \frac{10!}{5!(10-5)!} = 252 \] There are 252 different ways to select 5 problems from a set of 10 problems.
02

(Step 2: Find Ways to Correctly Answer All 5 Problems)

Since the student knows how to solve 7 of the problems, we can calculate the number of ways to select 5 problems from the 7 that the student knows how to solve as: \[ C(7,5) = \frac{7!}{5!(7-5)!} = 21 \] There are 21 ways for the student to correctly answer all 5 problems.
03

(Step 3: Calculate Probability of Correctly Answering All 5 Problems)

To find the probability of correctly answering all 5 problems, we must divide the number of ways the student can correctly answer all 5 problems (21) by the total number of ways to select 5 problems (252): \[ P(\text{Answer All 5 Correctly}) = \frac{21}{252} = \frac{7}{84} \approx 0.0833 \] The probability of the student correctly answering all 5 problems is approximately 0.0833 or 8.33%.
04

(Step 4: Find Ways to Correctly Answer At Least 4 Problems)

"At least 4 problems" means answering either 4 problems or 5 problems. We have already calculated the ways to answer all 5 problems correctly (21). Now, we will calculate the ways to correctly answer exactly 4 problems. The student knows how to do 7 problems, and 3problems are unknown. We can now find the combinations of 1. choosing 4 problems from the 7 known problems, 2. choosing 1 problem from the 3 unknown problems. This can be expressed as: \[ C(7,4) \cdot C(3,1) = \frac{7!}{4!(7-4)!} \cdot \frac{3!}{1!(3-1)!} = 35 \cdot 3 = 105 \] There are 105 ways for the student to correctly answer exactly 4 problems. Now we have to add the number of ways to answer 4 problems and 5 problems to find the ways to correctly answer at least 4 problems: \[ \text{Ways to Answer At Least 4 Problems} = 21 (\text{all 5}) + 105 (\text{exactly 4}) = 126 \]
05

(Step 5: Calculate Probability of Correctly Answering At Least 4 Problems)

To find the probability of correctly answering at least 4 problems, divide the number of ways to correctly answer at least 4 problems (126) by the total number of ways to select 5 problems (252): \[ P(\text{Answer At Least 4 Correctly}) = \frac{126}{252} = \frac{1}{2} = 0.5 \] The probability of the student correctly answering at least 4 problems is 0.5 or 50%. So, the final probabilities are: (a) The probability of correctly answering all 5 problems is approximately 8.33%. (b) The probability of correctly answering at least 4 problems is 50%.

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