Page 31 - Pra U STPM 2022 Penggal 3 - Maths (T)

Mathematics Semester 3 STPM Chapter 2 Probability

Let F and F be the events that the first and second printers developing fault
1
2
respectively.
We understand that events F and F are independent.
1
2
P(F ) = P(F ) = 0.2,
1
2
P(F ) = P(F ) = 1 – 0.2
2
1
= 0.8
Probability associated with each branch on the tree is written down.
From the tree diagram, two paths give the outcomes that one of the two printers
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
will develop a fault.
The probability that the first printer develops a fault, 2
P(F  F ) = P(F ) × P(F )
1
1
2
2
= 0.2 × 0.8
= 0.16
And the probability that the second printer develops a fault,
P(F   F ) = P(F ) × P(F )
1
2
1
2

= 0.8 × 0.2
= 0.16
Thus the probability that only one printer develops a fault
= P(F  F ) + P(F   F )
1
2
1
2
= 0.16 + 0.16
= 0.32.
Note: To use the tree diagram, multiply the probabilities along the branches of a path and add the
probabilities when more than one path fulfilling the requirements.
Example 35
A past record in a town provides the following information:
On a rainy day the probability of a driver involved in an accident is 0.08, whereas the probability of a
driver involved in an accident is 0.03 if there is no rain. The probability of rain in the town is forecasted
to be 0.25 in these few days. Find the probability that a driver will not involve in an accident tomorrow.
Solution: The tree diagram is displayed below where,
Event R: it is raining,
Event A: a driver involved in an accident.

Weather Accident or no accident
accident
0.08
rain
0.25
0.92 no accident
accident
0.03
0.75
no rain
0.97 no accident

99

02 STPM Math(T) T3.indd 99 28/10/2021 10:21 AM