Download Example. Lotto 6/49 Number of points in sample space 49 6

Example. Lotto 6/49
Number of points in sample space
49
6
!
= 13, 983, 816
You pick six numbers, say {1, 2, 3, 4, 5, 6}.
Let X =# of numbers you get right
x
0
1
2
3
4
5
6
P (X = x)
0.436
0.413
0.132
0.0177
0.000969
0.0000184
0.0000000715
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How are these obtained? Imagine a special deck of 6 red
& 43 black cards. The red cards are your six numbers.
The computer selects 6 cards at random and we count
how many of your numbers are matched. How do we get
the probability of matching five numbers, for instance.
Let A = {match five numbers}. Recall for equally likely
outcomes P (A) = |A|
. To match five, the computer
|S|
needs to match five out of your six numbers and one
out
of
43 numbers that you didn’t pick. There
are
6
43
ways to match five of six numbers and
5
1
ways to match 1 of 43 numbers. By the multiplication
rule, you
two terms together and
multiply
these
get
6
43
49
|A| =
×
. All together there are
5
1
6
numbers the computer could pick. So
6
43
49
P (A) =
5
1
6
2
More generally to match k numbers for k = 1, 2, 3, 4, 5, 6
6
43
49
49
P (X = 6) =
=1
6
0
6
6
6
43
49
P (X = 5) =
5
1
6
6
43
49
P (X = 4) =
4
2
6
6
43
49
P (X = 3) =
3
3
6
6
43
49
P (X = 2) =
2
4
6
6
43
49
P (X = 1) =
1
5
6
6
43
49
43
49
P (X = 0) =
=
0
6
6
6
6
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CONDITIONAL PROBABILITY
Probabilities do not live in a vacuum. They
are specified by conditions which may change
or about which additional information may become available. Consider A, B with P (A), P (B)
given. Suppose have knowledge that B occurred. How would the probability of A change?
Notation.
P (A|B)
denotes the
conditional probability of A given that B occurred.
This is a new probability.
defined?
How should it be
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Empirical Approach Industrial plant on a river
which forks downstream to two sites 1 and 2.
Let A denote the event that pollution is detected at site 1 on any day and B the event
that pollution is detected at site 2 on the same
day. What information is needed to specify
P (A|B)?
Method. Observe over n days the number of
days nB that site 2 is polluted and the number
of days nA∩B when site 1 is polluted. Then
take the ratio
nA∩B
nB
giving relative to pollution at site 2 the (relative) proportion of days when site 1 is polluted.
It is natural to call this the conditional probability of A given B.
5
But
nA∩B
proportion(A ∩ B)
n
/n
=
= A∩B
nB
nB /n
proportion(B)
and identifying proportion with probability we
get
Definition.
P (A|B) =
P (A ∩ B)
P (B)
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Example. Sometimes the experiment itself determines conditional probabilities. Shuffle deck
of cards and draw two cards at random. B =
{first card is an Ace}, A = {second card is an Ace }.
Natural to assert
3
P (A|B) =
51
Why? Now check using definition.
4
52
|A ∩ B|
4×3
P (A ∩ B) =
=
|B|
52 × 51
P (B) =
P (A|B) =
3
(4 × 3)/(52 × 51)
=
4/52
51
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Venn diagram representation of conditioning.
Conditioning reduces the sample space S.
..
This explains why previous example can be
solved in two ways.
Example. Components are assembled in plant.
Two different assembly lines, A and A0. On
a given day line A has assembled 80 components, of which 20 have been identified as defective (B) and 60 nondefective (B 0), whereas
A0 has produced 10 defective and 90 nondefective components.
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Item is selected at random.
80
= 0.44
180
Suppose item is checked and found to be defective. Reduce sample space.
P (A component) =
20
P (A component| defective) =
= 0.67
30
A
A0
B
20
10
B0
60
90
Multiplication Rule
Cross-multiply.
P (A ∩ B) = P (B)P (A|B)
Can be extended to more than 2 events.
For instance, for three events (see p76)
P (A1 ∩A2 ∩A3) = P (A3|A1 ∩A2)P (A2|A1)P (A1)
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Example 2.27 p76 Four individuals respond to
request for blood donations. Blood types unknown. Suppose only type O+ is desired and
only one of the four actually has this type. If
potential donors are selected at random, what
is probability that at least three individuals must
be typed to obtain O+?
Solution in text.
B = {first not O+}
A = {second not O+}
P (B) =
3
2
, P (A|B) =
4
3
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The multiplication rule now gives
P (at least three individuals needed) = P (A ∩ B)
= P (B)P (A|B)
3
2
=
= 0.5.
4
3
Alternate solution. Random permutation.
|S| = 4! = 24.
D = {O + third}, E = {O + fourth}
|D| = 3! = 6, |E| = 3! = 6,
6 + 6 = 0.5.
P (D ∪ E) = P (D) + P (E) = 24
24
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Example. Given P (A1) = 1/2, P (A2) = 1/3, P (A3) =
1/6, disjoint. Find P (A3|A2 ∪ A3).
P (A3 ∩ (A2 ∪ A3))
P (A3|A2 ∪ A3) =
P (A2 ∪ A3)
P (A3)
=
(why?)
P (A2) + P (A3)
1/6
=
= 1/3.
1/3 + 1/6
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