The document provides examples of using probability tree diagrams to calculate probabilities of independent events. It includes: 1) A tree diagram showing the possible outcomes of flipping two coins, with the probability of getting two heads (P(HH)) calculated as 1/4. 2) A tree diagram calculating the probability (0.38) that Helen has to stop exactly once at two sets of traffic lights, with stopping probabilities of 0.3 and 0.2. 3) A tree diagram calculating the probability (7/15) of picking one blue and one red sweet without replacement from a bag containing 3 blue and 7 red sweets.

1. Tree Diagrams
WALT complete a probability tree diagram for independent
events. Level 7
Starter
Draw a sample space diagram to show the possibilities
when you flip 2 coins.
What is P(2 Heads)?
Classwork
15/04/2020
Coin 1
Coin2
H T
H
T
HH HT
TH TT
𝟏
𝟒

2. I flip two coins, what is the probability that I get two heads?
Heads
Tails
Heads
Tails
Heads
Tails
1
2
1
2
1
2
1
2
1
2
1
2
P(HH) = x =1
2
1
2
1
4
P(HT) = x =1
2
1
2
1
4
P(TH) = x =1
2
1
2
1
4
P(TT) = x =1
2
1
2
1
4
Coin 1 Coin 2
Probabilities on
each branch add
up to 1 Possible outcomes are
shown on the end of the
branches
Multiply along the
branches to work out the
probabilities
P(HH) = 1
4

3. 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?
Stops
Doesn’t
Stop
Stops
Doesn’t
Stop
Stops
Doesn’t
Stop
3
10
0.7
0.8
0.2
0.8
0.2
P(SS) = 0.3 x 0.2 = 0.06
P(SD) = 0.3 x 0.8 = 0.24
P(DS) = 0.7 x 0.2 = 0.14
P(DD) = 0.7 x 0.8 = 0.56
Lights 1 Lights 2
= 0.24 + 0.14
= 0.38

4. Red
Blue
Blue
Blue
Red
Red
7
10
6
9
3
9
3
10
2
9
7
9
I have a bag of sweets, 3 blue and 7 red. I pick one sweet out without
looking, eat it and then pick out and eat a second sweet. What is the
probability that I eat one blue and one red sweet?
Sweet 1 Sweet 2
P(RR) = x =7
10
6
9
42
90
P(RB) = x =7
10
3
9
21
90
P(BR) = x =3
10
7
9
21
90
P(BB) = x =3
10
2
9
6
90
42
90
7
15
=