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Chapter 2: Problem 12

On a multiple-choice exam with three possible answers for each of the five questions, what is the probability that a student would get four or more correct answers just by guessing?

Short Answer

Expert verified
The probability of a student guessing and getting at least 4 correct answers is \(11/243\), or approximately 0.0452675.

Step by step solution

01

Understanding the binomial probability formula

The probability of getting exactly k successes in n independent Bernoulli trials with probability p of success on each trial can be calculated using the binomial probability formula: P(X = k) = \(\binom{n}{k}\) p^k (1-p)^(n-k) In this case, each question on the exam represents an independent Bernoulli trial with success probability p = 1/3 (since there are three possible answers, and only one correct answer). We are looking for the probability of getting at least four correct answers, so we will need to find P(X = 4) + P(X = 5).
02

Calculating the binomial coefficients

We will first need to compute the binomial coefficients \(\binom{n}{k}\) for the cases X = 4 and X = 5. These coefficients represent the number of ways we can choose k successes from n trials: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) For X = 4: \(\binom{5}{4} = \frac{5!}{4!(5-4)!} = 5\) For X = 5: \(\binom{5}{5} = \frac{5!}{5!(5-5)!} = 1\)
03

Computing the individual probabilities

Now, we will compute the probability of 4 and 5 correct answers using the binomial probability formula from Step 1: For X = 4: P(X = 4) = \(\binom{5}{4}\) (1/3)^4 (2/3)^1 = 5 * (1/81) * (2/3) = 10/243 For X = 5: P(X = 5) = \(\binom{5}{5}\) (1/3)^5 (2/3)^0 = 1 * (1/243) * 1 = 1/243
04

Computing the total probability

Finally, we can add the probabilities of getting 4 or 5 correct answers just by guessing: P(X >= 4) = P(X = 4) + P(X = 5) = 10/243 + 1/243 = 11/243 So, the probability of guessing and getting at least 4 correct answers is 11/243, or approximately 0.0452675.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{10}\) be independent Poisson random variable with mean \(1 .\) (i) Use the Markov inequality to get a bound on \(P\left[X_{1}+\cdots+X_{10} \geq 15\right]\). (ii) Use the central limit theorem to approximate \(P\left[X_{1}+\cdots+X_{10} \geq 15\right\\}\).

Suppose that \(X\) takes on each of the values \(1,2,3\) with probability \(\frac{1}{4} .\) What is the moment generating function? Derive \(E[X], E\left[X^{2}\right]\), and \(E\left[X^{3}\right]\) by differentiating the moment generating function and then compare the obtained result with a direct derivation of these moments.

In Exercise 2, if the coin is assumed fair, then, for \(n=2\), what are the probabilities associated with the values that \(X\) can take on?

Suppose that \(X\) and \(Y\) are independent binomial random variables with parameters \((n, p)\) and \((m, p) .\) Argue probabilistically (no computations necessary) that \(X+Y\) is binomial with parameters \((n+m, p)\).

An individual claims to have extrasensory perception (ESP). As a test, a fair coin is flipped ten times, and he is asked to predict in advance the outcome. Our individual gets seven out of ten correct. What is the probability he would have done at least this well if he had no ESP? (Explain why the relevant probability is \(P[X \geq 7\\}\) and not \(P(X=7) .\) )
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