Hi;
Not sure where I'm going wrong with this one.
Given that Susan fails just one exam, find the probability that its’ History she fails.
maths Pass = 0.7 french pass = 0.8 history pass 0.6
P(A|B) = P(A ∩ B) / P(B).
= (0.7 * 0.8 * 0.4) / ((0.7 * 0.8 * 0.4) + (0.7 * 0.2 * 0.4) + (0.3 * 0.2 * 0.4) + (0.3 * 0.8 * 0.4))
I get 0.56 book gets 0.4956
cant upload tree diagram sorry.
Thanks.
Hi Anthonye
maths Pass = 0.7 french pass = 0.8 history pass 0.6
Given that Susan fails just one exam, find the probability that its’ History she fails.
Probability that she fails History but passes both the others is 0.4*0.7*0.8 = 0.224 Just like you had.
Prob that she fails maths and passes the other 2 = 0.3*0.8*0.6 = 0.144
Prob that she fails franch and passes the other 2 = 0.2*0.7*0.6 = 0.084
The probability that she fails one and only one subject = 0.224+0.144+0.084 = 0.452
Given that Susan fails just one exam, find the probability that its’ History she fails = 0.224 / 0.452 = 0.4956 (4 dec places)
Another way of looking at it: writing "P" or "F" for pass or fail, and "Math", "French" and "History" in that order, there are possible outcomes, with probabilities:
PPP: (0.7)(0.8)(0.6)= 0.336
PPF: (0.7)(0.8)(0.4)= 0.224
PFP: (0.7)(0.2)(0.6)= 0.084
PFF: (0.7)(0.2)(0.4)= 0.056
FPP: (0.3)(0.8)(0.6)= 0.144
FPF: (0.3)(0.8)(0.4)= 0.096
FFP: (0.3)(0.2)(0.6)= 0.036
FFF: (0.3)(0.2)(0.4)= 0.024
Of those, PPF, PFP, and FPP correspond to "fails just one exam" and they have total probability 0.224+ 0.084+ 0.144= 0.452.
That one exam being French is PFP which has probability 0.084 so the probability that she will fail French given that she fails just one exam is