Probability (2)

(a) Four individuals have responded to a request by a blood bank for blood donations. None of them has donated before, so their blood types are unknown. Suppose only a specific type of blood is requested and only one of the four actually has this type. If the potential donors are selected successively and at random for blood typing, what is the probability that at least three individuals must be examined to obtain the desired type?

(b) Use the Multiplication Rule to prove that for three events A, B and C such that
P(A and B) > 0 and P(A) > 0,
P(A and B and C) = P(C|A and B)P(B|A)P(A)

I've done part (a) - let A1 = 1st individual not being desired type, and A2 = 2nd individual not being desired type,
so P(at least 3) = P(A1 and A2)
and P(A2 given A1) = P(A2 and A1)/P(A1)
so P(A2 and A1) = 2/3 * 3/4 = 1/2 = P(at least 3 tested)

how do I get started on proving part b? thanks

Reply 1

JohnSPals

stewiegriffin1

(b) Use the Multiplication Rule to prove that for three events A, B and C such that
P(A and B) > 0 and P(A) > 0,
P(A and B and C) = P(C|A and B)P(B|A)P(A)

Basically you keep applying the conditional probability formula, P(XnY)=P(X).P(Y|X), until you reduce the intersections.

So P(A n B n C)
= P((A n B) n C)
= P(A n B) . P(C|(AnB))
= P(A).P(B|A).P(C|(AnB))
As required.

You just have to see why you need that P(A and B) > 0 and P(A) > 0, which I'll leave to you.

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