5 Two students are supposed to take the final exam in CS 237

5. Two students are supposed to take the final exam in CS 237 but don\'t show up. They come to Snyder\'s office several hours after the exam and explain that they had to drive to visit a sick friend the morning of the exam, but got a flat tire on the way back, and missed the exam; they ask to be given a makeup exam. Snyder agrees. When they get the exam, it consists of two problems, the first (worth 5%) a simple problem from the course, and the second (worth 95%) consists of the single question \'\'Which tire was it?\'\' If the students have a 10% chance of actually having had a flat tire, then what is their chance of passing the exam? Assume that if they didn\'t have a flat tire, they have to guess randomly at the exam. Give the sample space in the case that they are lying, the appropriate event (i.e., the outcomes where they guess the same tire), the probability in the case that they are lying that they pass the exam, and finally the overall probability that they pass.

Solution

It is as if the problem from the course does not matter, because 95% of the exam is about the tire. So they would have to answer that part correctly.

If they really got a flat tire, they surely know it; hence, that\'s the 10% chance of flat tire.

If they didn\'t, they\'d have to guess.

Let the tires of their car be numbered 1,2,3,4.

Their possible guesses are (sample space)

1,1
1,2
1,3
1,4
2,1
2,2
2,3
2,4
3,1
3,2
3,3
3,4
4,1
4,2
4,3
4,4

Of which, only 4 of them matches. Hence, the probability of guessing right = 4/16 = 0.25.

So:

P(pass) = P(flat) + P(not flat) P(guessed right|not flat) = 0.10 + (1-0.10)(0.25) = 0.325 [ANSWER, overall probability of passing]

Also,

P(pass|not flat) = P(not flat) P(guessed right|not flat) = (1-0.10)(0.25) = 0.225 [ANSWER, lying but passed]