Page 27 - Pra U STPM 2022 Penggal 3 - Maths (T)

Mathematics Semester 3 STPM Chapter 2 Probability

(b) P(A) = 1 – P(A)
= 1 – 0.65
= 0.35
P(A  B) = P(B) – P(A  B)
= 0.3 – 0.15
= 0.15 S

P(B | A) = P(A  B) A B
P(A)
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
= 0.15 A B
0.35
= 3 2
7

Example 29

In a factory, three machines A, B and C are operated to make certain parts. The percentages of the parts
manufactured by the machines A, B and C are 35%, 50% and 15% respectively. It is known that 8%, 5%
and 16% of the parts produced by the machines A, B and C respectively are defective. If a finished part
is randomly picked, calculate the probability that the part is from the machine A given that it is defective.

Solution: Let event J = a part is made by machine A,
event K = a part is defective.
Assume that the total parts manufactured by the three machines are n. The
numbers of parts produced by machines A, B and C are 0.35n, 0.5n and 0.15n
respectively. Thus, the defective parts produced by machines A, B and C are
0.08 × 0.35n, 0.05 × 0.5n and 0.16 × 0.15n respectively.
Thus, P(K) = 0.08 × 0.35n + 0.05 × 0.5n + 0.16 × 0.15n
n
= 0.077
0.08 × 0.35n
and P(J  K) = n
= 0.028

Applying the conditional probability formula, the probability of the part made
by machine A given that the part is defective,
P(J | K) = P(J  K)
P(K)
= 0.028
0.077
= 0.364

Independent events

Suppose a fair coin is tossed and a ‘head’ is shown face up. What would the coin land on for the next toss?
The probability of getting a ‘head’ or a ‘tail’ is still 0.5. The outcome of the second toss is not affected by
the previous result. When the knowledge that an event has happened provides no information about the
occurrence of another event, the two events are said to be independent. Thus, if the outcome of event A
does not affect the outcome of event B, then A and B are independent events,
i.e. P(A | B) = P(A) or, equivalently P(B | A) = P(B)

95

02 STPM Math(T) T3.indd 95 28/10/2021 10:21 AM