Page 25 - Pra U STPM 2022 Penggal 3 - Maths (T)

Mathematics Semester 3 STPM Chapter 2 Probability
Hence, the conditional probability can be determined using the following formula.

P(A | B) = P(A  B) , provided that P(B) ≠ 0
P(B)

Rearranging the above equation gives
P(A  B) = P(A | B) × P(B)

Note: 1. P(B  A) = P(B | A) × P(A),
since P(A  B) = P(B  A)
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we deduce that, P(A | B) × P(B) = P(B | A) × P(A)
2. If A and B are mutually exclusive events, i.e. A  B = φ, then P(A  B) = 0. Thus, 2
P(A | B) × P(B) = P(B | A) × P(A) = 0.

3. For mutually exclusive and exhaustive events A and A, A  A = φ, A  A = S,
P(A  A | B) = P(A | B) + P(A | B)
i.e. P(S | B) = P(A | B) + P(A | B).
P(S | B) = P(S  B)
P(B)

= P(B)
P(B)
= 1
Hence, P(A | B) + P(A | B) = 1 or P(A | B) = 1 – P(A | B)
4. Consider the following Venn diagram:
S
A B
A B

A B
A = (A  B)  (A  B)
P(A) = P(A  B) + P(A  B)  (A  B) and (A  B) are mutually exclusive.
.
.
⇒ P(A) = P(A | B) P(B) + P(A | B) P(B)

Example 27

A bag contains 8 yellow and 4 green marbles. Two marbles are taken randomly from the bag. Determine
the probability that only one of them is yellow.

Solution: Let event A = first marble selected is yellow,
event B = second marble selected is yellow.

Then, P(A) = 8
12
= 2 .
3
When the second marble is selected, there are only 11 marbles left and 4 green
marbles remain untouched. So the conditional probability of B given that A has
happened is:

93

02 STPM Math(T) T3.indd 93 28/10/2021 10:21 AM