It's time for the finals
Five students are taking a final exam consisting of only a single multiple choice problem with four options. The testing is done online and the students are finishing at different times in a random order. Due to an error in the software, each student (except the first) is able to see the answer submitted earlier by one other, randomly selected, student. Only one of the students has studied and is confident. He will always stick to his answer, but his probability of making the correct choice is actually only . The remaining four are completely unprepared and will always copy.
What is the probability that the last person to finish gives the correct answer?
If the probability is in the form , where and are coprime positive integers, find .
The answer is 187.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
If the student with a clue happens to go first, probability of that happening is 1/5, his answer, which has 3/4 chance of being right, will get copied down, so the last student has a 3/4 chance of being right. Overall: 1/5 x 3/4 = 3/20.
If he is second, probability 1/5, the one who is third will get an equal chance to copy his answer or the answer of the first student, who had only 1/4 probability of being right. Average of those responses is 1/2, which is the probability of a correct answer. The one after him has again 1/2 chance of being right, averaging a 1/2 with what averaged to a 1/2, and so on. Overall: 1/5 x 1/2 = 1/10.
If he is third: average of 1/4, 1/4, and 3/4 becomes 5/12, giving overall probability of 1/5 x 5/12 = 1/12. Again unchanged after.
If he is fourth: average of 1/4, 1/4, 1/4, and 3/4 becomes 3/8, giving overall probability of 1/5 x 3/8 = 3/40.
If the student who studied is fifth, the probability of that happening being 1/5, he has his 3/4 chance of being right, giving overall probability of 1/5 x 3/4 = 3/10.
Combining the figures for the five alternatives: 3/20 + 1/10 + 1/12 + 3/40 + 3/ 20 = 67/120