Every Monday, James has a math class and a biology class.
The probability that he will have his math homework done is 0.46
(1) P(M) = P(M∩B) + P(M∩B') = 0.46
and the probability he will have his biology homework done is 0.57.
(2) P(B) = P(B∩M) + P(B∩M') = 0.57
If the probability he will have his biology homework done but not his math
homework is 0.21,
(3) P(B∩M') = 0.21
what is the probability he will have his math homework done given that he does
not have his biology done?
That's asking for
(4) P(M|B') = P(M∩B')/P(B')
Substitute (3) in (2)
P(B∩M) + P(B∩M') = 0.57
P(B∩M) + 0.21 = 0.57
(5) P(B∩M) = 0.36
Substitute (5) in (1), since P(M∩B) = P(B∩M) = 0.36
P(M∩B) + P(M∩B') = 0.46
0.36 + P(M∩B') = 0.46
(6) P(M∩B') = 0.10
(7) P(B') = 1-P(B) = 1-0.57 = 0.43
Substitute (6) and (7) in (4)
P(M|B') = P(M∩B')/P(B') = 0.10/0.43 = 10/43
Edwin
