1. Whiteboardmaths.com
© 2004 All rights reserved 5 7 2 1

2. + Probability: Independent Events 1 2 3 4 5 6 2 3 4 5 6 7 8 9 10 11 12
Consider the probability of throwing a double six with two dice. P(6 and 6)
We could make a table of all possible outcomes and count the ones that we want. 1 2 3 4 5 6 +
Complete the remainder of the table of outcomes. 2 3 4 5 6 7 8 9 10 11 12 3 4 4 5 5 6 6 7
P( 6 and 6) = 1/36 7 8

3. + Probability: Independent Events 1 2 3 4 5 6 7 8 9 10 11 12
The throwing of 2 dice are independent events. This means that the outcomes on one die are not affected in any way by the outcomes on the other. 1 2 3 4 5 6 + 7 8 9 10 11 12
On a single throw of a die P(6) = 1/6
For 2 dice the probability of 2 sixes is the same as 1/6 x 1/6 = 1/36
Can you see why multiplying the individual probabilities together for one event, (rolling a die) gives us the correct results for 2 events, (rolling 2 dice)?
This is because for each outcome on one die there are 6 outcomes on the other and so there are 6 x 6 = 36 outcomes in total.

4. 1 1 4 2 3 4 5 4 2 4 Spinners Probability: Independent Events
Consider the probability of getting a 2 on the blue spinner and a 4 on the red spinner.
Would multiplying the probabilities 2/5 x 3/6 = 6/30 give the correct answer? 1 1 2 3 4 4 2
Spinners 5 4 4

5. 1 2 2 + 1 4 5 3 4 Probability: Independent Events 1 2 3 4 5 3 2 5 6 4
Consider the probability of getting a 3 on the blue spinner and a 4 on the red spinner.
Probability: Independent Events
Would multiplying the probabilities 2/5 x 3/6 = 6/30 gives the correct answer? 2 3 1 4
Construct a table of outcomes to see whether or not this is correct. 4 2 5 1 1 2 3 4 5 + 3 2 5 6 4 7 8 9
A probability of 6/30 is correct.
Notice again that the events are independent of each other.

6. For independent events A and B, P(A and B) = P(A) x P(B)
Probability: Independent Events
AND LAW
The AND LAW
For independent events A and B, P(A and B) = P(A) x P(B)
P(6 AND 6)
P(3 BLUE AND 4 RED)
= 1/6 X 1/6 = 1/36
= 2/5 X 3/6 = 6/30
So rather than constructing tables of outcomes, or making lists of outcomes, we can simply multiply together the probabilities for the individual events.

7. For mutually exclusive events:
P(A and B) = P(A) x P(B)
The AND LAW
When using the And Law and multiplying the individual probabilities, the cumulative effect decreases the likelihood of the combined events happening. 3 2 1 1
Impossible
Certain
P(6) = 1/6
P(double 6) =
1/36
Common sense should tell you that it is much harder (less likely) to throw a double six with two dice, then it is to throw a single 6 with one die.
What do you think the probability of getting a triple 6 with a throw of 3 dice might be?
P(triple 6)
1/6 x 1/6 x 1/6 = 1/216

8. For mutually exclusive events:
The AND LAW
For mutually exclusive events:
P(A and B) = P(A) x P(B)
Impossible 1
Certain
When using the And Law and multiplying the individual probabilities, the cumulative effect decreases the likelihood of the combined events happening.
P(6) = 1/6
P(double 6) = 1/36 2 3
Common sense should tell you that it is much harder (less likely) to throw a double six with two dice, then it is to throw a single 6 with one die.
This reduction may seem obvious but it is worth stating explicitly since there is often confusion with the OR LAW (done earlier) where the probabilities increase.
Impossible 1
Certain
P(red) = 3/12
P(red or blue ) = 3/12 + 5/12 = 8/12
P(red or blue or black ) = 3/12 + 5/12 + 4/12 = 1 2 3
OR LAW

9. For independent events A and B, P(A and B) = P(A) x P(B)
The AND LAW
Probability: Independent Events
Both players lay a card at random from their hands as shown.
Question 1. What is the probability that two kings are laid?
Player 2
Player 1
P(king and king) =
2/7 x 1/7 = 2/49
Cards

10. For independent events A and B, P(A and B) = P(A) x P(B)
The AND LAW
Probability: Independent Events
Both players lay a card at random from their hands as shown.
Question 2. What is the probability that two hearts are laid?
Player 1
Player 2
P(heart and heart) =
4/6 x 2/8 = 8/48 (1/6)

11. The AND LAW Probability: Independent Events P(picture and picture) =
For independent events A and B, P(A and B) = P(A) x P(B)
The AND LAW
Probability: Independent Events
Both players lay a card at random from their hands as shown.
Q 3. What is the probability that two picture cards are laid? (Ace is not a picture card)
Player 1
Player 2
P(picture and picture) =
2/8 x 5/7 = 10/56 (5/28)

12. Beads (a) P(red and red) = 4/9 x 4/9 = 16/81 (b) P(blue and red) =
Question 4. Rebecca has nine coloured beads in a bag. Four of the beads are red and the rest are blue. She removes a bead at random from the bag and notes the colour before replacing it. She then chooses a second bead.
Calculate the probability that Rebecca chooses:
(a) 2 red beads (b) A blue followed by a red bead.
Beads
(a) P(red and red) =
4/9 x 4/9 = 16/81
(b) P(blue and red) =
5/9 x 4/9 = 20/81

13. (a) P(red and red) = 3/8 x 6/9 = 18/72 = 1/4 (b) P(blue and blue) =
Question 5. Peter and Rebecca each have a bag of red and blue beads as shown below. They each remove a bead at random from their bags.
Peter selects his bead first.
Calculate the probability that:
(a) Both beads will be red (b) Both beads will be blue
(c) Peter’s bead is red and Rebecca’s is blue. (answers in simplest form)
(a) P(red and red) =
3/8 x 6/9 = 18/72 = 1/4
(b) P(blue and blue) =
5/8 x 3/9 = 15/72 = 5/24
(c) P(red and blue) =
3/8 x 3/9 = 9/72 = 1/8

14. 5 1 2 2 3 4 5 2 2 4 Spinners Probability: Independent Events
The pointers on both spinners shown below are spun.
Calculate the following probabilities:
(a) A 2 on both spinners. (b) A 2 on the blue spinner and a 5 on the red spinner. 5 1 2 3 4 2 2 5 2 4
(a) P(2 and 2) =
2/5 x 3/6 = 6/30 (1/5)
(b) P(2 and 5) =
2/5 x 2/6 = 4/30 (2/15)

15. 6 5 4 8 4 3 Probability: Independent Events (a) P(4 and 4) =
The pointers on both spinners shown below are spun.
Calculate the following probabilities:
(a) A 4 on both spinners. (b) A 5 on the red spinner and an 8 on the blue spinner. 3 4 8 4 6 5
(a) P(4 and 4) =
2/6 x 3/8 = 6/48 (1/8)
(b) P(5 and 8) =
3/6 x 3/8 = 9/48 (3/16)

16. 2 4 6 9 3 8 1 5 Probability: Independent Events (a) P(4 and 5) =
The pointers on the spinners are spun in the order shown.
Calculate the probabilities for the outcomes shown on the following pairs of spinners :
(a) Pink and red (b) Pink and blue (c) Red and blue 4 1 2 3 5 6 8 9
(a) P(4 and 5) =
2/5 x 3/6 = 6/30 (1/5)
(b) P(4 and 6) =
2/5 x 3/8 = 6/40 (3/20)
(c) P(5 and 6) =
3/6 x 3/8 = 9/48 (3/16)

17. Worksheet