1. No. 1 c) 7 20 d) 3 40 Leaving work Getting to work on time on time5
on time
on time 7 8
P(TL) = x = 7 8 2 5 7 20 2 5
late 3 5
on time 1 8
Mutually exclusive branches 3 5
P(LT) = x = 1 8 3 40
late 2 5
late
Probabilities on each branch add up to 1
Multiply along the branches to work out the probabilities
THE SAME PAIR OF BRANCHES

2. No. 2 d) 0.64 c) 0.04 Second Pin First Pin 0.2 0.2 0.8 Down 0.2 0.8UP
P(both Up) = 0.2 x 0.2 = 0.04
0.2 UP
0.8
Down
0.2 UP
0.8
Mutually exclusive branches
Down
P(both down) = 0.8 x 0.8 = 0.64
0.8
Probabilities on each branch add up to 1
Down
Multiply along the branches to work out the probabilities
THE SAME PAIR OF BRANCHES

3. No. 3 c) 2 d) 2 5 15 Try No. 4 Second Box First Box lime white yellow
P(WL) = x = 2 5 1 3 2 5 2 15
lime
white 1 3
Try No. 4 3 5
yellow 2 5
lime 2 3
Mutually exclusive branches
yellow 3 5 2 3 3 5 2 5
P(Both Y) = x =
yellow
Probabilities on each branch add up to 1
Multiply along the branches to work out the probabilities
THE SAME PAIR OF BRANCHES

4. No. 4 a) 0.35 c) 0.15 b) 0.15 d) 0.35 Crossley Fires Becker Fires HIT
P(both Hit) = 0.5 x 0.7 = 0.35
Becker Fires
HIT
0.7
HIT
0.5
P(B,:C) = 0.5 x 0.3 = 0.15
0.3
MISS
0.7
P(:B,C) = 0.5 x 0.7 = 0.35
HIT
0.5
Mutually exclusive branches
MISS
0.3
MISS
Probabilities on each branch add up to 1
P(Both don’t Hit) = x 0.3 = 0.15
THE SAME PAIR OF BRANCHES

5. Remember
When we move along branches to find the probability of this AND this we multiply the individual probabilities!
For independent events: the two sets of branches on the second level are the same
BECAUSE
No matter what happens in the first event – the possible outcomes of the second event will be the same

6. What is the probability that she has to stop exactly once? Discuss
Helen passes through 2 sets of traffic lights on her way to work.
The probability that she stops at the first set is 0.3
The probability that she stops at the second set is 0.2
What is the probability that she has to stop exactly once? Discuss
Lights 2
Lights 1
Stops
P(SS) = 0.3 x 0.2 =
0.06
0.2
Stops
Doesn’t
Stop
0.3
P(SD) = 0.3 x 0.8 =
0.24
0.8 OR
Stops
P(DS) = 0.7 x 0.2 = 0.14
0.2
0.7
Doesn’t
Stop
0.8
Doesn’t
Stop
P(DD) = 0.7 x 0.8 = 0.56
P(exactly once) = = 0.38

7. mutually exclusive events
AND / OR
This AND This are independent events
we multiply the probabilities of independent events
AND = MULTIPLY
(This AND This) OR (That AND That) are
mutually exclusive events
we add the probabilities of mutually exclusive events
OR = ADD

8. L.O. To be able to find any combined probability from a probability tree diagram
By the end of the lesson we will be able to:
identify when a question is asking us to multiply along the branches (AND)
identify when a question is asking us to consider more than one path (OR)
understand what to do when a question has the words ‘at least’
answer any probability question set about a probability tree diagram