Page 26 - Pra U STPM 2022 Penggal 3 - Maths (T)

Mathematics Semester 3 STPM Chapter 2 Probability

P(B | A) = 4 . By applying the conditional probability formula,
11
P(yellow first and green second) = P(A  B) = P(B | A) × P(A)
= 4 × 2
11 3
8
= 33
Similarly, P(A) = 4 = 1 .
12 3
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When the second marble is selected, there are only 11 marbles left and 8 yellow
marbles are still there. The conditional probability of B given that A has occurred
8
2 is thus determined and P(B | A) is equal to 11 .
Using the conditional probability formula,
P(green first and yellow second) = P(A  B) = P(B | A) × P(A)
= 8 × 1
11 3
= 8 .
33
P(only one yellow) = P(first yellow followed by second green or first green
followed by second yellow) = P(A  B) + P(A  B)
= 8 + 8
33 33
= 16 .
33
The probability that only one of the two marbles taken is yellow is 16 .
33
Note: The result whether the two marbles are taken from the bag at the same time or one after the other
is the same.

Example 28

A and B are two events in a sample space S such that:
P(A | B) = 0.5, P(A) = 0.65 and P(B) = 0.7. Calculate
(a) P(B | A), (b) P(B | A).

Solution: (a) P(B) = 1 – P(B)
= 1 – 0.7
= 0.3
P(A | B) = P(A  B)
P(B)
0.5 = P(A  B)
0.3
P(A  B) = 0.15
P(B | A) = P(A  B)

P(A)
= 0.15
0.65
= 3
13

94

02 STPM Math(T) T3.indd 94 28/10/2021 10:21 AM