Download File - Mathematics with Mr Walters

STATISTICS S1
SLBS
PROBABILITY
Relative frequency
=
Number of times a particular event occurs
Total number of events
Sample space is a list of all possible outcomes
Ex 5A p 79
VENN DIAGRAMS
A
6
4
8
B
3
The values inside the Venn Diagram maybe numbers (of items)
or probabilities (as decimals or fractions)
A ∩ B is the INTERSECTION
A  B is the UNION
A/ is the COMPLEMENT of A (i.e. NOT A)
P(A') = 1 – P(A)
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STATISTICS S1
P(AB) =
SLBS
4
21
A
6
4
8
B
3
P(AB) =
18
21
A
6
4
8
B
3
P(A') =
11
21
A
6
4
8
B
3
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STATISTICS S1
P(A'B) =
SLBS
8
21
A
6
4
8
B
3
P(AB)' =
3
21
=
1
7
A
6
4
8
B
3
Ex 5B p 83
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STATISTICS S1
SLBS
ADDITION RULE
P(AB) = P(A) + P(B) – P(AB)
Example
P (A) = 0.4, P (B) = 0.7,
Find P (A ∩ B)
P (A  B) = 0.9
0.9 = 0.4 + 0.7 − P (A ∩ B)
P (A ∩ B) = 0.4 + 0.7 − 0.9 = 0.2
Example
The probability of event A occurring is 0.4
The probability of event B occurring is 0.7
The probability that neither A nor B occurs is 0.2
Find the probability that both A and B occur.
A
0.4-x
x
0.7-x
B
0.4
0.7
0.2
(0.4 − x) + x + (0.7 − x) + 0.2 = 1
1.3 − x
= 1
x
= 0.3
Therefore, P (A ∩ B) = 0.3
Hence all regions could now be completed
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STATISTICS S1
SLBS
A
0.1
0.3
0.4
B
0.2
Ex 5C p 86
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STATISTICS S1
SLBS
THE MULTIPLICATION RULE
P (A│B) represents the probability that A occurs given that B
has occurred.
It is known as a CONDITIONAL PROBABILITY
A
6
4
8
B
3
P(AB) =
Hence
Note also:
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4
12
=
1
3
=
4/21
12/21
=
P(AB)
P(B)
P(AB) = P(AB)  P(B)
P(AB) = P(BA)  P(A) = P(AB)  P(B)
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STATISTICS S1
SLBS
There are times when you may have to use a triple Venn
diagram.
Example
In the Sixth form, 60% of the students study Mathematics,
40% study Physics and 30% study Chemistry. 20% study
Mathematics and Physics, 22% study Physics and Chemistry,
18% study Maths and Chemistry, whilst 13% study all three
subjects and 17% do not study any of these three subjects.
Represent this information on a Venn diagram and use it to find
the probability that a student selected at random:
a) is studying Maths only.
b) is studying Maths and Physics but not Chemistry
c) is studying Maths given that the student is also studying
Physics
d) is studying Physics, given that, the student is also
studying Maths and Chemistry.
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STATISTICS S1
SLBS
100
60
M
40
20
7
35
13
5
P
11
9
18
22
3
C
30
17
a) P(M only) = 35% = 0·35
b) P(M & P not C) = 7% = 0·07
c) P(MP) =
P(MP)
d) P(PM&C) =
P(P)
=
20
40
P(PMC)
P(MC)
= 0·5
=
13
18
= 72·2%
Ex 5D p 91
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STATISTICS S1
SLBS
TREE DIAGRAMS
Example
A fair coin is tossed twice.
Find the probability of obtaining:
a)
two Heads
b)
one Head and one Tail
c)
at least one Head
d)
the second coin lands Tails given that the first coin
lands Heads
0.5
H
HH = 0.5 x 0.5 = 0.25
T
HT = 0.5 x 0.5 = 0.25
H
TH = 0.5 x 0.5 = 0.25
T
TT = 0.5 x 0.5 = 0. 25
H
0.5
0.5
0.5
0.5
T
0.5
a)
P(HH) = 0.5 x 0.5 = 0.25
b)
P(IH & 1T) = P(HT or TH) = 0.25 + 0.25 = 0.5
c)
P(at least 1H) = 1 − P(No Heads) = 1 − 0.25 = 0.75
d)
P(T2│H1) = P(T2∩H1)/ P(H1) =
0·25
= 0.5
0·5
Note: this may be read directly off the tree diagram
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STATISTICS S1
SLBS
Sometimes, conditional probabilities cannot be read directly
off the tree diagram
Example
Three coins are such that one of them is biased and he other
two are fair. The probability that the biased coin lands Heads
up is 0.7. If one of the coins is tossed twice and lands Heads
up on both occasions, what is the probability that it is the
biased coin?
The two events here are choosing the coin and tossing the coin
twice.
P (2H on fair coin) = 0.5 x 0.5 = 0.25
P (2H on biased coin) = 0.7 x 0.7 = 0.49
2H
Biased
B & 2H = 1/3 x 0.49 = 0.16333...
Not 2H
2H
F & 2H = 2/3 x 0.25 = 0.16666.....
Fair
Not 2H
P (2H) = 0.16333… + 0.16666… = 0.33
P (Biased│2H)
= P (Biased ∩ 2H) / P (2H)
0·16333
=
= 0·495 to 3 s.f.
0·33
Ex 5E p 94
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STATISTICS S1
SLBS
INDEPENDENT & MUTUALLY EXCLUSIVE EVENTS
Read pp 81-82
Two events are INDEPENDENT if the occurrence (or nonoccurrence) of one event has no effect on the occurrence (or
non-occurrence) of the other event
P (A│B) = P (A)
P(AB) = P(A)  P(B)
Therefore,
Two events are MUTUALLY EXCLUSIVE if they cannot happen
simultaneously
i.e. A or B but not both
P(AB) = 0
A
B
P(AB) = P(A) + P(B)
Two events are MUTUALLY EXHAUSTIVE if they cover all
possibilities.
A
B
P(AB) = 1
0
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STATISTICS S1
SLBS
Example
A and B are mutually exhaustive events
1
1
P(A'B) =
P(AB') =
3
6
Are A and B independent, mutually exclusive or neither?
B
A
1/6
x
1/3
0
1
6
+ x +
P(AB) =
1
3
1
2
= 1  x =
1
2
 A and B are NOT mutually exclusive
[Also P(AB) = 1  P(A) + P(B) =
P(A)  P(B) =
2
3

5
6
=
5
9
2
3
+
5
6
]
 P(AB)  A and B are NOT independent
Read Examples 13-16 p 96
Ex 5F p 100
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STATISTICS S1
SLBS
NUMBER OF ARRANGEMENTS
What’s In A Name!
Read pp 88-90
No.of arrangements =
e.g.
n!
p!  q!  r!  ...
No. of arrangements of AABBB =
5!
2!  3!
= 10
Example
A bag contains 5 Red counters and 4 Blue counters. Three are
selected at random without replacement. Find the probability
that 2 are Red and 1 is Blue.
(You could use a Tree diagram, or……)
P(R1 R2 B3) =
5
9

4
8

4
7
No. of arrangements of RRB =
P(2 Red and 1 Blue) = 3 
5
9

3!
2!
4
8
= 3

4
7
=
10
21
Mixed Ex 5G p 104
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