PB Triangle Area

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P(B) Triangle Area
P(A) Oval Area
P(A and B) w
P(A or B) r w g
So,
P(A) P(B)
r w g w
The Addition Rule for 2 events A and B P(A or B)
P(A) P(B) - P( A and B)
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What is the probability of choosing a red tile
at random?
2/9
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Event Five blue tiles are removed
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What is the probability now, given the event?
1/2
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Conditional Probability
A
B
A?B
A
P(AB) P(A and B)/ P(B)
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• An Example
• Suppose 5 of 2m cars in a community are bad
  (B), 20 are brand A and 2 are both bad and
  brand A (A and B).
• Question What is the probability that a car is
  bad, given that it is brand A?
• Answer A 400000 B 100000
• B and A 40000
• P(BA) P(B and A)/P(A)
• (40000/2m)/(400000/2m) 0.1

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• The probability of event B is dependent on the
  the event A.
• A and B are not independent events.

Statistical independence Two events e and f are
statistically independent if and only if
P(ef) P(e)
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P (e f) P(e and f)/P( f)

• The Multiplication rule
• For two events e and f
• P (e and f) P (e f) P ( f)

If e and f are independent then
P (e and f ) P (e) P( f)
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An example to show that two events can be
mutually exclusive and dependent
A bag contains 6 black and 4 red balls.
Experiment Two balls are chosen one after the
other (with replacement)
A Event the first ball is red
B Event both are black
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P(A) 4/10 0.4
P(B) 6/10 6/10 0.36
P(A and B) 0
So A and B are mutually exclusive
But,
P(BA) P(A and B)/P(A) 0
So, P(BA) ? P(B)
A and B are not independent events
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An example to show that two events can be
independent yet not mutually exclusive
A community has 1m people.
600,000 are women (W)
500,000 are adults (A)
300,000 adults are women (A and W)
Obviously A and W are not mutually exclusive
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P(AW) P(A and W)/P(W) 300,000/600,000 0.5
Also, P(A) 500,000/1 million 0.5
Thus, P(AW) P(A)
A and W are independent events
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• Arrangements and Selections
• Consider the following two problems
•  a. How many ways two letters can be selected
  from the set containing the three, A, B and C,
  (order of selection important)
• b.  How many ways two letters can be selected
  from the set containing the three, A, B and C,
  (order of selection not important)

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A
B
19
A
C
20
B
A
21
B
C
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C
A
23
C
B
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• In problem a, the answer is AB, AC, BA, BC, CA,
  CB.
• The problem is relevant in choosing a password,
  for example.
• The password AB123 is not the same as BA123 !
• Each of the six possibilities is called an
  arrangement or a permutation.
•  

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• The 3 represents the number of letters to choose
  from and the number of terms in the
  multiplication represents the number of letters
  to choose.

Altogether, we have 6 3 X 2 ways
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• In problem b, the answer is AB, BC and CA, a
  total of 3 different selections or combinations.
• This problem is relevant when selecting members
  of a committee, for example.
• The committee consisting of member A being
  selected first and then B is not different from
  the one in which B was chosen first and then A.

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Arrangements
Selections
AB
BA
AB
CA
AC
AC
BC
CB
BC

• Now consider problem c
• How many ways three letters can be selected from
  the set containing the three, A, B, C. (order of
  selection important)?

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C
A
B
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B
A
C
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A
B
C
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C
B
A
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B
C
A
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A
C
B
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ABC
BAC
BCA
ACB
CAB
CBA
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• The 3 represents the number of letters to choose
  from and the number of terms in the
  multiplication represents the number of letters
  to choose.

Altogether, we have 6 3 X 2 X 1 ways
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• In general, if r letters are to be selected
  from n possible ones, where (r n), order of
  selection important, the total number of
  selections is
• nPr nx(n-1)x(n-2)x(n-r1)
• If r n, we have the total number of selections
  nx(n-1)x(n-2).x2x1 n! (read factorial n).
• We cannot choose the same answer for problem b as
  that of a, since order is unimportant.

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• AB is no different from BA!
• So for every two selections in a, we need to take
  only one. Hence the answer is 3.

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Arrangements
Selections
AB
BA
AB
CA
AC
AC
BC
CB
BC
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• In general, if r letters are to be selected
  from n possible ones, where (r n), order of
  selection not important, we first start with the
  number
• nPr nx(n-1)x(n-2)x(n-r1)
•  Then we realise that the r letters selected can
  themselves be arranged in r! ways.
• Since order is not important, we should just
  count only one selection for every r! of them.

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• So the required number is
•  nCr nPr /r! nx(n-1)x(n-2)x(n-r1)/rx(r-1
  )x(r-2)x x2x1
• Thus, the number of 5-member committees formed
  out of 20 possible candidates is
• 20C5 (20 x 19 x 18 x 17 x 16) / (5 x 4 x 3 x
  2 x 1) 15504
• Examples

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• 1. A committee of three is to be selected from 3
  boys and 4 girls. It must have exactly 1 boy and
  2 girls in it. How many such committees can be
  formed?
• Let the boys be called B1, B2 and B3, the girls
  G1, G2, G3 and G4 First, consider those
  selections in which B1 is selected. So two girls
  have to chosen to form the committee.

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• Two girls out of four can be chosen in 4C2
  4x3/2x1 6 ways. So we shall have 6 committees
  in which B1 is a member.
• But what about committees of which (i) B2 is a
  member and (ii) B3 is a member?
• There are six of each type above.
• So there are 6 x 3 18 committees altogether.

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• The formal way of answering the question is as
  follows
• 1 boy can be chosen out of 3 in 3C1 3
  possibilities.
• For each such selection of a boy, 2 girls out of
  4 can be selected in 4C2 w 6 different ways.
• So the total number of possible selections is 3C1
  X 4C2 18

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• 2. The UK National Lottery
• (i) What is the probability of winning the
  jackpot from a single ticket?
• (ii) What is the probability of winning exactly
  10 from a single ticket?
• Answer (i) If 6 balls are chosen out of 49 then
  there are 49C6 or 13983816 ways of doing so.

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• The machine can select any one of 13983816
  combinations.
• So P(one ticket wins jackpot)
• 1/ 13983816
• Answer (ii) Consider the six (yet unknown)
  numbers to be chosen by the machine and call them
  winners.
• So there are 6 winners and 43 losers.
• A ticket wins 10 if it selects exactly 3 winners
  out of 6 and 3 losers out of 43.