A stats prof has noted from past experience that students who do the homework for the course have a 0.90 probability of passing the course. On the other hand, students who do not do the homework for the course have a 0.25 probability of passing the course. The prof estimates 75% of the students in the course do the homework. Given a student passes the course, what is the probability that she or he completed the homework?
2007-10-10 17:19:28 · 3 answers · asked by a.pasternak 2 in Science & Mathematics ➔ Mathematics
H - event that student does HW
P - event that student passes
RTF: P(H|P) = P(H and P) / P(P)
P(H) = 0.75
P(P|H) = 0.9
P(P|H') = 0.25
P(P) = P(P|H)*P(H) + P(P|H')*P(H')
= 0.9*0.75 + 0.25*0.25
= 0.7375
P(H and P) = P(P|H)*P(H)
= 0.9*0.75 = 0.675
P(H|P) = 0.675/0.7375
= 54/59
2007-10-10 17:26:36 · answer #1 · answered by Dr D 7 · 1⤊ 0⤋
For computational sake, take 100 students. The problem indicates that 75 do homework and 25 dont. Of the 75, 90% or 67.5 students will pass.
Of the 25 slackers, 25% or 6.25 student will pass.
Thus, 73.75 students of 100 are expected to pass. So the probability of a passing student having done homework is 67.5/73.75
2007-10-11 00:33:17 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
P=0.9*0.75/(0.9*0.75+0.25*0.25)
=2.7/(2.7+0.25)
=2.7/2.95
=54/59
2007-10-11 12:49:39 · answer #3 · answered by Mugen is Strong 7 · 0⤊ 0⤋