00:02
So here we're given the following problem, which is asking on different events involving the blood types o, a, b, and ab.
00:11
And we're given the following distributions or the probability distribution for 0, a, b, and ab, the probability of each blood type.
00:21
And also we need to know that o can donate to anybody.
00:32
A can donate to a and a, b, b can donate to b, b can donate to v.
00:40
B and ab and ab can only donate to itself.
00:49
All right, so now let's look at the first question that's being asked.
00:54
Consider the following events.
00:57
Also we know that patient a or patient one has type a and patient two is type b.
01:03
So let's also write that.
01:05
I'm going to write p1, a, p2, b.
01:11
All right, so the first event is d1, d1, a randomly chosen person can be a donor for patient one.
01:20
D2, a randomly chosen person can be a donor for patient two.
01:24
So first we are interested in find the probability that a randomly chosen person can be a donor for patient one or patient two.
01:30
So we're finding pd1 or pd2.
01:33
First we need to find pd1 and pd2.
01:36
Okay, so pd1 is a probability that a person or that patient one is able to receive a transplant.
01:47
Or a donation.
01:49
And we know that p1 is type a, which means that it can receive a donation only if the donor is type o or type a.
02:00
And so this is simply equal to the probability that someone is type o or that someone is type a or plus, and because these are mutually exclusive events, you can't be type o and type a.
02:14
You just add the individual probabilities because this case, the mutually disjoint rule works.
02:22
You get 0 .44 plus 0 .42 equals 0 .86...