I'm doing this problem:

A student is applying to Harvard and Dartmouth. He estimates that he has
a probability of .5 of being accepted at Dartmouth and .3 of being accepted
at Harvard. He further estimates the probability that he will be accepted by
both is .2. What is the probability that he is accepted by Dartmouth if he is
accepted by Harvard? Is the event “accepted at Harvard” independent of the
event “accepted at Dartmouth”?

So I begin with the p(D|H)=p(D and H)/p(H) which I find eerily suspicious because is too easy, so is wrong, but why? First the probabilities do not sum to one p(H)+p(D)≠1. So I'm missing something obvious but what??


  • $\begingroup$ This surely seems like homework. If so, you should add the self-study tag. $\endgroup$ – Peter Flom Nov 21 '13 at 1:11
  • $\begingroup$ Why should p(H) + p(D) = 1? $\endgroup$ – Peter Flom Nov 21 '13 at 1:12
  • $\begingroup$ Tag added, because those are the only 2 options I see, maybe p(no college) is .2?? $\endgroup$ – Pedro.Alonso Nov 21 '13 at 1:23
  • 1
    $\begingroup$ P(H) + P(no H) = 1. So does P(D) + P(no D). However, here there are 4 possibilities: H only, D only, H+D, and neither. Those 4 together have to = 1. $\endgroup$ – Peter Flom Nov 21 '13 at 1:26
  • $\begingroup$ Ha ok I see it, I've forgotten H', ok then I will try something. Thanks. $\endgroup$ – Pedro.Alonso Nov 21 '13 at 1:28

You should be able to draw and use something like the diagram below to lay out the information (write it in the spaces and margins) and then you may be able to see how to do the problem. You would write all of the values on the diagram and from them fill in probabilities for every subregion and colored region(/margin). Then you should have a clearer idea what everything is.

diagram for conditional probability problem

Since (from his post in chat) the OP has worked the last details out correctly; here's my outline of the answer:

Given the values in the question (in blue), we can infer the other probabilities (in dark red) by subtraction:

diagram with values filled in

a) The "probability that he is accepted by Dartmouth if he is accepted by Harvard" $= P(D|H) = P(D\cap H)/P(H) = 0.2/0.3 = 2/3$

b) "Is the event “accepted at Harvard” independent of the event “accepted at Dartmouth”?".

The two events are independent if $P(D\cap H)=P(D)P(H)$; equivalently they are independent if $P(D|H)=P(D)$ (as long as $P(H)$ is not 0).

In the first approach $P(D)P(H)= 0.5\times 0.3 = 0.15 \neq P(D\cap H)=0.2$, while using the second approach, $P(D|H)=2/3\neq P(D)=0.5$, so either way the events are not independent.

  • $\begingroup$ I have done it like that, I get p(H')=0.5, p(H)=0.5, p(D)=0.3, p(D')=0.7, p(D and H)=0.2, p(D and H')=0.1, p(D' and H)=0.3, p(D' and H')=0.4, and p((H)*p(D)=0.15 therefore they are not independent. Is this correct? $\endgroup$ – Pedro.Alonso Nov 21 '13 at 15:12
  • $\begingroup$ Not quite (unless I made an error). Check your answers for p(D and H') and p(D' and H). But you're on the right track. $\endgroup$ – Glen_b Nov 21 '13 at 23:21
  • $\begingroup$ Your conclusion about independence is right, and looks like it's for the right reason. $\endgroup$ – Glen_b Nov 21 '13 at 23:39
  • $\begingroup$ Well I cant find the mistake, let me go tru this again p(H)=0.5, then p(H')=0.5, p(D)=0.3, p(D')=0.7, correct? Then in the table of the joint probabilities I have subtracted the marginal and the one joint I have like: p(D and H')=p(D)-p(D and H), 0.3-0.2=0.1, correct? $\endgroup$ – Pedro.Alonso Nov 22 '13 at 2:11
  • $\begingroup$ And summing the 4 entries gives me 1, 0.1+0.4+0.2+0.3=1, so why is this? Can you tell me the mistake it will be better. :) $\endgroup$ – Pedro.Alonso Nov 22 '13 at 2:26

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