Page 37 - DMTH404_STATISTICS

Unit 3: Conditional Probability and Independence Baye’s Theorem

Notes
Example 13: A manufacturer of automobile parts knows from past experience that the
probability that an order will be completed on time is 0.75. The probability that an order is
completed and delivered on time is 0.60. Can ygu help him to find the probability that an order
will be delivered on time given that it is completed ?
Let A be the event that an order is delivered on time and H the event that it is completed on time.
Then P(H) = 0.75 and P(A  H) = 0.60. We need P(A | H).

P(A  H) 0.60
P(A|H)    0.8.
P(H) 0.75
Have you understood the definition of conditional probability? You can find out for yourself by
doing these simple exercises.

Task If A is the event that a person suffers from high blood pressure and B is the
event that he is a smoker, explain in words what the following probabilities represent.
(a) P(A | B)
(b) P(Ac | B)
c
(c) P(A | B )
(d) P(Ac | Bc).
Two unbiased dice are rolled. They both show the same score. What is the probability that
their common score is 6?

We now state some of the properties of P(A | H).

P1 : For any setA,  P(A | H )  1.
Recall that since A  H  H, P(A  H)  P(H). The required property follows immediately.
P2 : P(A | H) = 0 if and only if A  H is a null set. In particular, P( | H) = 0 and P(A | H) = 0 if A
and H are disjoint events.
P3 : P(A | H) = 1 if and only if P(A  H) = P(H).
In particular,
P( | H)= 1 P(H | H) = 1

P’4: P(A  B | H) = P(A | H) + P(B | H) – P(A  B | H).
How do we get P’4 ? Well, since
(A  B)  H = (A  H)  (B  H),
P2 gives us

P((A  B)  H) = P(A  H) + P(B  H) – P(A  B  H).
Now use the definition of the conditional probability to obtain P4.
Using P’4 and P3 and P4 of Sec. 6.2.2, we get
P5 : If A and B are disjoint events, P(AB | H) = P(A  H) + P(B | H)

C
and P(A | H) = 1 – P(A | H)

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