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"John is taking an algebra and a calculus class. The probability that he will pass algebra is 0.57 and the probability he passes calculus is 0.68. The probability he passes calculus given he passes algebra is 0.72, and the probability that he passes calculus given he doesn't pass algebra is 0.45. What is the probability he passes algebra given he passes calculus."

My approach:

$$P(C|A)\cdot P(A) = P(A \cap C)$$

Then $\dfrac{P(A\cap C)}{P(C)} = 0.6035$,

but this is incorrect according to the professor, and that the answer should be 0.6796

Can anyone explain this? Thank you in advance!

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Below are the detailed calculation. Then reason why I provide all the details is because it seems to be a mistake in the statement.


Apply Bayes' formula to obtain

$$ \Pr(A \,|\, C) = \frac{\Pr(C \,|\, A) \Pr(A)}{\Pr(C)} $$

The numerator is $0.72 * 0.57 = 0.4104$.

The denominator can be calculated using the law of total probability

$$ \Pr(C) = \Pr(C \,|\, A) \Pr(A) + \Pr(C \,|\, \text{not }A) \Pr(\text{not }A) $$ $$ = 0.72 * 0.57 + 0.45 * 0.43 = 0.6039 $$

Thus,

$$ \Pr(A \,|\, C) = \frac{0.4104}{0.6039} = 0.6796 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{(not 0.6769)} $$


Note that the statement says that $\Pr(C) = 0.68$, which, according to the above calculation, is wrong.

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  • $\begingroup$ you beat me by 38 seconds!!! +1 $\endgroup$ – MikeP Sep 30 '16 at 14:21
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    $\begingroup$ ;-) (I will +1 you too ;-)) $\endgroup$ – ocram Sep 30 '16 at 14:23
  • $\begingroup$ Thank you. The professor said that "the probability he passes calculus is 0.68" should have not been included. Thank you for your help! $\endgroup$ – mikedaabeast Sep 30 '16 at 21:18
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I think the critical factor is to ignore the given that the probability he passes calculus is 68%. This is actually not correct given the other information in the problem.

If the probability that he passes algebra is 57%, the probability that he passes calculus given passing algebra is 72%, and the probability that he passes calculus while failing algebra is 45%, then $$P(C) = P(C|A)P(A) + P(C|A')P(A') = 0.57*0.72 + (1-0.57)*0.45 \approx 0.604$$

Dividing $P(C \cup A)$ by 0.6039 instead of 0.68 will give you the 0.68 the prof is looking for.

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Expand the denominator using Law of Total Probability and the correct answer follows.

Your mistake was to write $P(C|A)\cdot P(A) = P(A \cap C)$ since this is also equal to $P(A|C)\cdot P(C)P(A|C)\cdot P(C)$ where $P(A|C)P(A|C)$ is the quantity we want to find, so this would become a self-referential problem and hence unsolvable this way.

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  • $\begingroup$ Can you point out his mistake? $\endgroup$ – Memming Sep 30 '16 at 13:46

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