S1 Mutually exclusive and independent events

Report
MUTUALLY EXCLUSIVE
AND INDEPENDENT EVENTS
•To understand mutually exclusive events
•To understand independent events
•To understand when to use the addition and
multiplication rules
•To be able to draw venn diagrams for different types
of events
When events have no outcomes in common they are mutually exclusive.
P(A  B) = 0
Using the addition rule
P(AB) = P(A) + P(B) - P(A  B)
Since P(A  B) = 0
Then P(AB) = P(A) + P(B) (The OR rule)
E.G. Getting a 2 or a 3 in one roll of a dice.
A
B
1
6
1
6
S
4
6
When one event has no effect on another they are independent events.
The probability of B happening is the same whether A has happened or not.
P(A  B) = P(A) x P(B) (The AND rule)
E.G. Getting a 5 on a spinner numbered 1,2,3,4,5,5,7,8 and a 3 on a fair
unbiased dice.
A
B
S
A red die and a blue die are rolled and the outcome of each die is
recorded.
A=outcome on red die is 3
B=outcome on blue die is 3
C=sum of the two dice is 5
D=the outcome on each die is the same
Show that A and B are independent
P(A)= 6 = 1
Red die
1
B
l
u
e
d
i
e
1
2
3
4
2
3
36
4
5
6
6
P(B)= 6 = 1
36
6
P(A  B) = 1 x 1 = 1
6
6
36
5
This is the same as can be seen on
the sample space diagram
6
Therefore A and B are independent
A red die and a blue die are rolled and the outcome of each die is
recorded.
A=outcome on red die is 3
B=outcome on blue die is 3
C=sum of the two dice is 5
D=the outcome on each die is the same
Show that C and D are mutually exclusive
Red die
1
B
l
u
e
d
i
e
1
2
3
4
5
6
2
3
P(C) has 4 outcomes
4
5
6
P(D) has 6 outcomes
They have no overlapping
outcomes in common.
Therefore C and D are mutually
exclusive
A bag contains 7 green beads and 5 blue beads. A bead is taken from the
bag at random, the colour is recorded and the bead is replaced. A
second bead is then taken from the bag and its colour is recorded.
a) Find the probability that one bead is green and the other is blue
b) Show that the event “the first bead is green” and “the second bead is
green” are independent
7
12
7
12
G
5
12
B
5
12
7
12
5
12
G P(one bead is green and the
other is blue)
=P(GB) + P(BG)
B =7 x 5 + 5 x 7
12 12 12 12
G
= 35 + 35
144
144
B = 70 = 35
144 72
A bag contains 7 green beads and 5 blue beads. A bead is taken from the
bag at random, the colour is recorded and the bead is replaced. A
second bead is then taken from the bag and its colour is recorded.
a) Find the probability that one bead is green and the other is blue
b) Show that the event “the first bead is green” and “the second bead is
green” are independent
P(first is G) = 7
12
7
G
P(second is G) =P(GG) + P(BG)
12
7
12
5
12
G
5
12
7
12
=7 x 7 + 5 x 7
B 12 12 12 12
5
12
P(GG) = 7 x 7
B
12 12
This is the same as multiplying P(G) x
P(G) on the tree diagrams, therefore the
events are independent.
B
G = 49 + 35
144
= 84 = 7
144 144 12

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