Find probability of someone who has cheated on the test and solved a question within 20% of allocated time.

[poojashinde]

To check who are the students that cheat on a test, a college tries to find relations. Given the following information, what is the probability that someone who has finished the test within 20% of allocated time has cheated on the test?
1) 1% students cheat on the test.
2) 20% students solve a question within 20% allocated time.
3) 80% students that cheat on a test solves the questions in within 20% of allocated time.

My solution: Consider 500 students. Thus, 5 cheat, 200 solve in 20% of allocated time, 80% of people that cheat(i.e. out of 5) 4 of them also solve in 20% of allocated time. Thus out of 500- 4 students cheat and solve it in 20% of allocated time i.e. 4/5 out of 100. So my guess is 0.8 should be the probability.
But I'm not sure. Can someone help me ?

 

[JeffM]

Student is a cheater abbreviated as C. P(C) = 0.01.

Student is quick solver abbreviated as Q. P(Q) = 0.2.

Student is a quick solver given that student is a cheater. P(Q | C) = 0.8

Question, what is the probability that a student is a cheater given that student is quick. P(C | Q) = what?

[MATH]\text {P(C | Q)} = \dfrac{\text {P(C and Q)}}{\text {P(Q)}} = \dfrac{\text {P(Q | C)} * \text {P(C)}}{\text {P(Q)}} = \dfrac{0.8 * 0.01}{0.2} = 4\%. [/MATH]
Now that is the formulaic way. Here is the arithmetic way. Out of 500 students,
500 * 0.01 = 5 cheat and
500 * 0.2 = 100 finish quickly (not 200 but I think you meant 100)
5 * 0.8 = 4 cheaters who finish quickly.

Therefore 4/100 of those who finish quickly cheat.