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7 2
1 5
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Probability: Independent Events
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.
Complete the remainder of
the table of outcomes.
+
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10 11
6
7
8
9
10 11 12
P( 6 and 6) = 1/36
Probability: Independent Events
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.
On a single throw of a die P(6) = 1/6
+
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10 11
6
7
8
9
10 11 12
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.
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 2
4
3
2
1
4
2
Spinners
4
5
4
Probability: Independent Events
Consider the probability of getting a 3 on the blue spinner and a 4 on the red spinner.
Would multiplying the probabilities 2/5 x 3/6 = 6/30 gives the correct answer?
Construct a table of outcomes to see
whether or not this is correct.
2 3
4 +1
1
3
2
1
2
4
4
4
5
2
3
5
5
5
6
3
4
6
6
6
7
3
4
5
7
7
7
8
3
4
5
7
7
7
8
4
5
6
8
8
8
9
4
5
1
4
2
4
A probability of
6/30 is correct.
Notice again that the events are independent of each other.
Probability: Independent Events
The AND LAW
AND LAW
For independent events A and B, P(A and B) = P(A) x P(B)
P(6 AND 6)
= 1/6 X 1/6 = 1/36
P(3 BLUE AND 4 RED)
= 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.
For mutually exclusive events:
The AND LAW
P(A and B) = P(A) x P(B)
When using the And Law and multiplying the individual probabilities, the
cumulative effect decreases the likelihood of the combined events happening.
2
3
1
½
0
1
Certain
Impossible
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
For mutually exclusive events:
The AND LAW
P(A and B) = P(A) x P(B)
When using the And Law and multiplying the individual probabilities, the
cumulative effect decreases the likelihood of the combined events happening.
2
3
1
½
0
1
Certain
Impossible
P(double 6) = 1/36
P(6) = 1/6
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.
1
0
½
2
3
OR LAW
1
Certain
Impossible
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
The AND LAW
Probability: Independent Events
For independent events A and B, P(A and B) = P(A) x P(B)
Both players lay a card at random from their hands as shown.
Question 1. What is the probability that two kings are laid?
Player 1
P(king and king) =
Player 2
2/7 x 1/7 = 2/49
Cards
The AND LAW
Probability: Independent Events
For independent events A and B, P(A and B) = P(A) x P(B)
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
P(heart and heart) =
Player 2
4/6 x 2/8 = 8/48 (1/6)
The AND LAW
Probability: Independent Events
For independent events A and B, P(A and B) = P(A) x P(B)
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
P(picture and picture) =
Player 2
2/8 x 5/7 = 10/56 (5/28)
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.
(a) P(red and red) =
(b) P(blue and red) =
4/9 x 4/9 = 16/81
5/9 x 4/9 = 20/81
Beads
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
Probability: Independent Events
The pointers on both spinners shown below are spun.
Spinners
Calculate the following probabilities:
(a) A 2 on both spinners. (b) A 2 on the blue spinner and a 5 on the red spinner.
1 2
4
3
2
2
5
5
4
2
2
(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)
Probability: Independent Events
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.
6
5
4
4
5
5
3
8 4
8
4
4 38
(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)
Probability: Independent Events
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
4 2
4
3
1
6
5
(c) Red and blue
4
5
5
8
3
5 6
9
4
6 36
(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)
Worksheet