S1: Probability
Chapter 3
E: Tree diagrams
It is very useful to use when:
• One event is repeated or
• More than one event occurs
Example 1:
Two
Late events
The probability that Jeffrey is late for college on
On time
any one day is 0.15 and is independent of whether he was late on the
previous day. Find the probability that Jeffrey is:
A. Late on Monday and Tuesday
Monday
Tuesday
Late
B. Arrives on time on one of these
days.
Outcome
Probability
Late, Late
0.15 x 0.15 = 0.0225
Late, On time
0.15 x 0.85 = 0.1275
On time, Late
0.85 x 0.15 = 0.1275
Late
On
time
Late
On
A – Read from
time tree
On
time
On time, On time
0.85 x 0.85 = 0.7225
Therefore: Probability = 0.1275 + 0.1275 = 0.255
NOTE:
Where outcomes are not independent, the
conditional probabilities are written on the branches:
OUTCOME
P( A  B)  P( A)  P( B | A)
Example 2:
A trainee has two tests, A and B. The probability
of passing A is 0.8. If the trainee passes A, the probability of passing B
is 0.9. If they fail A, the probability of passing B is 0.6.
A.
Find the probability of
passing B.
2 Events: Pass and Fail
(0.8  0.9)  (0.2  0.6)  0.84
P(pass
at O
least
1) = 0.92
METH
D 2
First draw a tree diagram:
2. P(Pass A and B) + P(Pass A, Fail B) +
P (Fail A, Pass B)
P( passA  passB) 0.0
8
0.9
.72
C. Given that a trainee passes Pass
B, B
From A:
find
the probability
that of
they
pass A.
B. Find
the probability
passing
at
Pass A
least one of the tests.
Fail B P(passB) = 0.84
2
M EAT H O D S Pass B
Fail
1. P(pass at least 1) P
= (1passA
– P (fail
 both)
passB)
P( passA | passB) = 1 – (0.2 X 0.4)
Fail B
P( passB)
= 0.92
0.72
P( passA | passB) 
0.84
= 0.857
(3sf)
P( passB)  P( passA & passB)  P( failA & passB)
REMEMBER: 27 Jan Tuesday
Probability test first lesson
Exercise E page 55
Numbers 1 to 6
Remember to try (on website):
Practise exam questions sheet