# Thread: Conditional Probability/Sequence of Events

1. ## Conditional Probability/Sequence of Events

The four major bloodtypes in the US are A (42%), B (10%), AB (4%) and O (44%) where the percentages denote the proportion of the population.

If four people are picked at random, let P(k) be the chance that there are exactly k different bloodtypes among the four people. What is the probability P(k) for k = 1,2,3,4?

Now theoretically I can do this, but it would take me half an hour. Is there a simpler way of thinking about this? P(1) is easy to compute, so maybe there is a way to build on that to get P(2) and so on. Any help?

2. Originally Posted by mbaboy
The four major bloodtypes in the US are A (42%), B (10%), AB (4%) and O (44%) where the percentages denote the proportion of the population.

If four people are picked at random, let P(k) be the chance that there are exactly k different bloodtypes among the four people. What is the probability P(k) for k = 1,2,3,4?

Now theoretically I can do this, but it would take me half an hour. Is there a simpler way of thinking about this? P(1) is easy to compute, so maybe there is a way to build on that to get P(2) and so on. Any help?
Hi mbaboy,

To me, cases k = 1, 4, and 3 seem the easier ones, with k = 3 being a little more complicated than 1 and 4. So I would try those cases and then use

P(1) + P(2) + P(3) + P(4) = 1

to find P(2).

Hint: Are you familiar with the multinomial distribution?

3. I can find P(1) and P(4), but P(3) and P(2) are too complicated to compute. I've never heard of multinomial distribution. I'm one week into an introductory probabily class. All we have done are iterations of Bayes' Theorem.

4. Originally Posted by mbaboy
I can find P(1) and P(4), but P(3) and P(2) are too complicated to compute. I've never heard of multinomial distribution. I'm one week into an introductory probabily class. All we have done are iterations of Bayes' Theorem.
OK, here's the idea for computing P(3). If there are 3 blood types then there must be one blood type of which you have 0, two of which you have 1, and one of which you have 2 in your sample. There are 12 such possibilities in all. For example, you might have N(AB) = 0, N(A) = 2, N(B) = 1, N(O) = 1. Work out the probability in each of the 12 cases and add the results to get P(3). The natural tool for this computation is the multinomial distribution (to me), but I think you can do it with conditional probability without mucn difficulty.

Hope this helps...