1. Baumgartner, Prosser, and Crowell are grading a calculus exam. There is a true-false question with ten parts. Baumgartner notices that one student has only two out of the ten correct and remarks, “The student was not even bright enough to have flipped a coin to determine his answers.” “Not so clear,” says Prosser. “With 340 students I bet that if they all flipped coins to determine their answers there would be at least one exam with two or fewer answers correct.” Crowell says, “I’m with Prosser. In fact, I bet that we should expect at least one exam in which no answer is correct if everyone is just guessing.” Who is right in all of this?

This problem is a Bernoulli trial and can be solved with Theorem 3.6.

Theorem 3.6 Given n Bernoulli trials with probability \(p\) of success on each experiment, the probability of exactly \(j\) successes is

\[b(n, p, j) = \bigg ( \begin{matrix} n\\ j \end{matrix} \bigg ) p^jq^{n-j} \\ where \space q = 1 - p.\]

First let’s look Baumgartner’s implied assumption that if the students all flipped coins there will be no exam with two or fewer answers correct.

\[n = 340\]

To calculate \(p\) we must sum the possible permutations of getting 0 success of 10 trials, 1 success of 10 trials, and 2 successes of 10 trials, and divide this sum by the total possible subsets from 10 elements.

\[ p = \frac{\bigg ( \begin{matrix} 10\\ 0 \end{matrix} \bigg ) + \bigg ( \begin{matrix} 10\\ 1 \end{matrix} \bigg ) + \bigg ( \begin{matrix} 10\\ 2 \end{matrix} \bigg )}{2^{10}} = \frac{56}{1024}\] \[j = 0\]

\[b(340, \frac{56}{1024}, 0) = \bigg ( \begin{matrix} 340\\ 0 \end{matrix} \bigg ) \frac{56}{1024}^0\frac{968}{1024}^{340} = 4.961765e-09\]

It seems that it is almost improbable that there will no exams with two or fewer correct answers if all 340 students flipped a coin. So Baumgartner is wrong.

Now let’s look at Crowell’s claim that there will be atleast one exam with all incorrect answers.

To reduce the computational workload with factorials, I will use the opposite of this claim by saying that there will be no exam with zero correct answers if they all guessed. If there is a high probability of getting zero correct answers then it would not be reasonable to expect atleast 1 exam with zero correct answers.

\[n = 340\]

To calculate \(p\) we must find the possible permutations of getting zero success of ten trials and divide it by the total possible subsets from 10 elements.

\[ p = \frac{\bigg ( \begin{matrix} 10\\ 0 \end{matrix} \bigg ) }{2^{10}} = \frac{1}{1024}\]

\[j = 0\]

\[b(340, \frac{1}{1024}, 0) = \bigg ( \begin{matrix} 340\\ 0 \end{matrix} \bigg ) \frac{1}{1024}^0\frac{1023}{1024}^{340} = 0.7173485\]

It seems that Crowell is wrong to expect at least one exam with all incorrect answers. Ofcourse this depends on your threshold for expectation.

Only Prosser is correct in expecting at least 1 one exam with two or fewer correct answers if they all guessed!