Human beings have 4 types of basic blood type: AB, A, B and O. In the US, about 43% of people are O type, 43% of people are A type, 10% of people are B type and 4% of people are AB type1. Now you are a doctor and have two patients waiting for blood transfusion. One of them has blood type O and can only accept blood type O transfusion; the other has blood type A and can receive either type O or A transfusion. Assume every donor’s contribution can only support one patient (that is, one donor cannot donate blood to both patients). If you choose 5 donors randomly from the population, what’s the probability that both of the patients are able to get blood transfusions?
In order for the patients to get blood transfusions, we need to get at least 2 O-types or an O-type and an A-type, and the rest of the three can be anything. Thus, our probabilities are: \begin{align} \mathbb{P}(getting\;O,O)&=\mathbb{P}(getting;O)\cdot\mathbb{P}(getting\;O)\cdot\mathbb{P}(anything)\cdot\mathbb{P}(anything)\cdot\mathbb{P}(anything)\\ &=(0.43)(0.43)(1)(1)(1) \end{align} and \begin{align} \mathbb{P}(getting\;O,A)&=\mathbb{P}(getting;O)\cdot\mathbb{P}(getting\;A)\cdot\mathbb{P}(anything)\cdot\mathbb{P}(anything)\cdot\mathbb{P}(anything)\\ &=(0.43)(0.43)(1)(1)(1) \end{align} We also need to find the probability of not getting OO or AO. \begin{align} \mathbb{P}(not\;getting\;O,O)&=\mathbb{P}(not\;O)^4\cdot\mathbb{P}(anything)\\ &=(0.57)^4(1) \end{align} and \begin{align} \mathbb{P}(not\;getting\;O,A)&=\mathbb{P}(not\;O\;and\;not\;A)^3\cdot\mathbb{P}(anything)\cdot\mathbb{P}(not\;A\;or\;not\;O)\\ &=(0.14)^3(1)(0.57) \end{align} so \begin{align} \mathbb{P}(not\;O,O\;and\;not\;O,A)=(0.57)^4(1)\cdot(0.14)^3(1)(0.57) \end{align}
I think I should use a multinomial distribution to represent the events. Let X=event that we get OO, Y=event that we get OA, and Z=event that we get neither. Then $$(X,Y,Z)\sim\text{Multinom}(n,r,p_1,p_2,p_3)$$. But I am not sure what parameters to use. Is this the right approach to the problem?
Strictly going by the question, the only way at least one of the patients does not get blood transfusion is when
i) one of the donors is blood group $O$ and rest four are either $B$ or $AB \ $ OR
ii) none of the donors are blood group $O$.
So probability that at least one of the patients does not get blood transfusion $ = 5 \times 0.43 \times 0.14^4 + 0.57^5 \approx 0.061 $
So probability that both of them get blood transfusion $ \approx 0.939$.