# Thread: conditional probability problem

1. ## conditional probability problem

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.

3. ## Re: conditional probability problem

manage to upload tree diagram.

4. ## Re: conditional probability problem

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)

5. ## Re: conditional probability problem

Thanks Melody2 I've sorted it out. these type of questions seem to be written the
wrong way round to me ie P(B) is what were given but its the first part of the question.

6. ## Re: conditional probability problem

Yes, getting your head around these problems is often the hardest part.
i am glad that you have it sorted.

7. ## Re: conditional probability problem

Thanks again for all your help.

8. ## Re: conditional probability problem

Another way of looking at it: writing "P" or "F" for pass or fail, and "Math", "French" and "History" in that order, there are $\displaystyle 2^3$ 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 $\displaystyle \dfrac{0.084}{0.452}= 0.1858$

9. ## Re: conditional probability problem Originally Posted by HallsofIvy Another way of looking at it: writing "P" or "F" for pass or fail, and "Math", "French" and "History" in that order, there are $\displaystyle 2^3$ 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 $\displaystyle \dfrac{0.084}{0.452}= 0.1858$
Ooops! I misread the problem! It was "History", not "French". That will be PPF which has probability $\displaystyle \dfrac{0.224}{0.452}= 0.4956$, the same as Melody2 said. Whew!

10. ## Re: conditional probability problem

Hi HallsofIvy,
I was a bit confused when your name came up.
Nice to meet you again!

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