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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 is 1/42.

Step by step solution

01

Calculate the total possible outcomes

To find the total possible outcomes, we need to find the total number of ways to pick 5 questions from a list of 10. This can be calculated using combinations. In general, the number of combinations (C) of choosing k items out of n is given by: C(n, k) = n! / (k!(n-k)!) In this case, we have n = 10 and k = 5: C(10, 5) = 10! / (5!(10-5)!) = 252 So, there are 252 possible outcomes of choosing 5 questions from the list of 10.
02

Calculate the favourable outcomes

Now, we need to find the number of favourable outcomes, which are combinations where all 5 selected questions are among those the student can solve. We know the student can solve 6 problems. So, this would be the number of ways to choose 5 problems out of the 6 that the student can solve. Using the same combination formula with n = 6 and k = 5: C(6, 5) = 6! / (5!(6-5)!) = 6 There are 6 favourable outcomes where the student can solve all 5 problems on the exam.
03

Calculate the probability

To find the probability that the student can solve all 5 problems on the exam, we will divide the favourable outcomes by the total possible outcomes: Probability = Favourable Outcomes / Total Possible Outcomes = 6 / 252 = 1 / 42 The probability that the student can solve all 5 problems on the exam is 1/42.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinations
When tackling problems involving combinations, it's essential to understand that combinations are a way to calculate how many different groups or subsets you can form from a larger set. Unlike permutations, combinations don't take into account the order of items, which makes combinations incredibly useful in probability.
  • The combination formula is given by: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] Here, \( n \) represents the total number of items to choose from, and \( k \) represents how many items you want to choose.
  • "!" denotes a factorial, which means multiplying a series of descending natural numbers. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
For example, in this exercise, you're selecting 5 questions from a total of 10. Using combinations, you determine there are 252 possible ways or groups in which this selection can be done.
Favourable Outcomes
In probability, favourable outcomes refer to the specific instances or cases that satisfy the condition for which you're calculating the probability. Identifying these outcomes correctly is key to solving probability problems.
  • In our problem, the favourable outcomes are the number of ways in which the student can be given problems she can solve.
  • She's able to solve 6 questions out of the 10. Thus, we calculate how many groups of 5 questions can be formed solely from the 6 she knows. This ensures each group (or outcome) is favourable.
The combination formula here tells us there are 6 such groups. Thus, there are 6 favourable outcomes for this problem scenario.
Probability Calculation
Once you have the number of favourable outcomes and the total possible outcomes, the next step is calculating the probability of the desired event. Probability is essentially the fraction of favourable outcomes over the total possible outcomes.
  • The formula for probability is: \[ \text{Probability} = \frac{\text{Favourable Outcomes}}{\text{Total Possible Outcomes}} \]
  • This fraction represents how likely it is that a certain event will happen.
For instance, the probability that our student can solve all 5 questions presented to her is calculated as: \[ \frac{6}{252} = \frac{1}{42} \] This translates to a 1 in 42 chance.

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Most popular questions from this chapter

Three students are playing a card game. They decide to choose the first person to play by each selecting a card from the 52 -card deck and looking for the highest card in value and suit. They rank the suits from lowest to highest: clubs, diamonds, hearts, and spades. a. If the card is replaced in the deck after each student chooses, how many possible configurations of the three choices are possible? b. How many configurations are there in which each student picks a different card? c. What is the probability that all three students pick exactly the same card? d. What is the probability that all three students pick different cards?

A random variable \(x\) has this probability distribution: $$\begin{array}{l|llllll}x & 0 & 1 & 2 & 3 & 4 & 5 \\\\\hline p(x) & .1 & .3 & .4 & .1 & ? & .05\end{array}$$ a. Find \(p(4)\). b. Construct a probability histogram to describe \(p(x)\). c. Find \(\mu, \sigma^{2},\) and \(\sigma\). d. Locate the interval \(\mu \pm 2 \sigma\) on the \(x\) -axis of the histogram. What is the probability that \(x\) will fall into this interval? e. If you were to select a very large number of values of \(x\) from the population, would most fall into the interval \(\mu \pm 2 \sigma\) ? Explain.

A study is to be conducted in a hospital to determine the attitudes of nurses toward various administrative procedures. If a sample of 10 nurses is to be selected from a total of 90 , how many different samples can be selected? (HINT: Is order important in determining the makeup of the sample to be selected for the survey?)

How many times should a coin be tossed to obtain a probability equal to or greater than .9 of observing at least one head?

Suppose a group of research proposals was evaluated by a panel of experts to decide whether or not they were worthy of funding. When these same proposals were submitted to a second independent panel of experts, the decision to fund was reversed in \(30 \%\) of the cases. If the probability that a proposal is judged worthy of funding by the first panel is \(.2,\) what are the probabilities of these events? a. A worthy proposal is approved by both panels. b. A worthy proposal is disapproved by both panels. c. A worthy proposal is approved by one panel.
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