My girlfriend asked for my help on a probability problem, and I'm kind of stumped. I don't want to let her down so I was hoping I could get a tip. This is the question:

Suppose we have two hypothesis $H_1$ and $H_2$ with disjoint probabilities such that $P(H_1)=0.2$ and $P(H_2)=0.8$ Now we observe an event $E$, which would be unlikely given $H_2$. Specifically, the probability of seeing event $E$ given $H_2$ is 40%, $P(E | H_2)= .4$.

How can I calculate the conditional probabilities $P(H_1|E)$ and $P(H_2|E)$$?

I know Bayes theorem is:

$$P(A | B) = (P(B | A) * P(A)) / P(B)$$

And when I try to apply this to the question at hand:

$$ P(H2 | E) = \frac{P(E | H_2) * P(H_2)}{P(E)} \\ =\frac{ 0.4 * 0.8 }{ P(E) }\\ = \frac{.32 }{P(E)} $$

However, $P(E)$ is not given. Similarly when I attempt to use Bayes formula for the other conditional probability:

$$ P(H_1 | E) = \frac{P(E | H_1) * P(H_1) }{ P(E) }\\ = \frac{P(E | H_1) * 0.2 }{ P(E)} $$

Again, I do not know $P(E)$, let alone $P(E | H_1)$.

If I try to use the information I already have and Bayes formula to derive some new information, I solve for nothing useful:

$$ P(E | H_2) = \frac{P(H_2 | E) * P(E)}{ P(H_2)} = 0.4\\ = \frac{P(H_2 | E) * P(E) }{ 0.8 } = 0.4\\ => P(H_2 | E) * P(E) = .32 $$

Is is possible that I am missing some information needed to solve the problem?

I appreciate any help, thanks!

  • $\begingroup$ There is missing information, as you say. But let me ask you this. What is the exact phrasing of the question she's having trouble with? Because if the question states "what are the conditional probabilities of each hypothesis given the event?" then you have not enough information. But if the question asks "which hypothesis is more likely given observation of the event?" then, indeed, you do have enough information. $\endgroup$ Aug 27, 2017 at 23:46
  • $\begingroup$ The question asks to "calculate the conditional probabilities and determine which is the more likely hypothesis". $\endgroup$
    – Nick vP
    Aug 28, 2017 at 0:01

1 Answer 1


To the question of what the exact value the posterior probabilities take, there is missing information. More specifically, there is one piece of information missing. You only need $P(E \mid H1).$ You could also get $P(E)$ and that would be enough as well. The reason you only need one of them is because you could infer one from the other using the sum rule of probability, $$ P(E) = P(E \mid H1) P(H1) + P(E \mid H2) P(H2). $$

However, for the question, "Which hypothesis is more likely given $E$," you actually do have enough information. To see this, look at the ratio of posterior probabilities of each hypothesis.

$$ \frac{P(H1 \mid E)}{P(H2 \mid E)} = \frac{P(E \mid H1) P(H1)} {P(E \mid H2) P(H2)} = \frac{1}{4} \frac{P(E \mid H1)} {0.4}. $$ The posterior probability for $H1$ is greater if the ratio above is greater than one. Now, what condition does $P(E \mid H1)$ have to satisfy in order for the above ratio to be greater than one?


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