Section 2.7 Conditional Probability

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Chapter 2, Section 7AConditional Probability
? John J Currano, 09/17/2007
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Game Toss a die and observe the top face Sample
Space S 1, 2, 3, 4, 5, 6 You win if the
event A 1, 2, 3 occurs.


)
(
win)
(
A
P
P
Now suppose that the die has been tossed and
while you are not told what face shows, you are
told the number is odd i.e., that the event B
1, 3, 5 has occurred. Would you still say
P(win) 0.5? Equivalently, would both players
still be willing to bet even money?
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Game Toss a die and observe the top face Revised
Sample Space B 1, 3, 5 You win if the event
A ? B 1, 3 occurs.
Recall that in the unconditional case (no
additional knowledge) Sample Space S 1, 2,
3, 4, 5, 6 You win if the event A 1, 2, 3
occurs.
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The conditional probability formula that we have
derived works only for finite sample spaces in
which all outcomes are equally likely (a uniform
discrete probability model).
Dividing both numerator and denominator of the
fraction by S does not change its value and
gives us a formula which we use for the
definition of conditional probability
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Definition The conditional probability of the
event A given the event B is
provided P(B) ? 0.
We do not define P(AB) when P(B) 0.
Example. Toss a fair coin three times and note
the resulting sequence of Heads and Tails. 1.
S

• 2 ? 2 ? 2 8
• Let A be the event the first toss is H, and
• find P(A) A/S

.
H A 1 ? 2 ? 2
4/8 1/2
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• Example. Toss fair coin 3 times note sequence of
  Heads Tails.
• S 8
• 2. Let A be the event the first toss is H P(A)
  A/S 4/8.
• 3. Let B be the event at least two heads show
  find P(AB).
• We can find P(AB) in two ways since S is uniform
  discrete

Since we know neither P(B) nor P(A ? B), we need
to calculate B and A ? B anyway, so well use
the second fraction. To calculate B and A ?
B, we can either list the sample points in each,
or try to use our counting techniques.
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Example. Toss fair coin 3 times note sequence of
Heads Tails. A first toss is H B at least
two heads show 1. S 8 2. P(A) A/S
4/8 3. FindP(A B). Method 1 List the
elements of A ? B and B.
A ? B HHT, HTH, HHH A ? B 3
A ? B
H .
B HHT, HTH, THH, HHH B 4
B
.
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Example. Toss fair coin 3 times note sequence of
Heads Tails. A first toss is H B at least
two heads show 1. S 8 2. P(A) A/S
4/8 3. FindP(A B). Method 2 Use counting
techniques to find A ? B and B.
Elements of A ? B have H followed by at least one
more H so choose either 1 or 2 of the remaining
blanks for the other Hs
.
H
Elements of B have at least two Hs so choose
either 2 or 3 of the blanks for the Hs
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.
Example. A bowl contains 7 blue chips and 3 red
chips. Two chips are drawn at random, in order
and without replacement. Find the probability
that the first is red and the second is
blue. There are at least two ways to approach
this problem. 1. Use permutations or the
multiplication rule. 2. Use conditional
probability.
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.
Bowl has 7 blue chips and 3 red chips. Two are
drawn at random, in order w/out replacement.
Find P(first is red second is blue). 1. Use the
multiplication rule. Even though the chips of a
given color are indistinguishable from each
other, we pretend that they can be told apart
e.g., pretend each has a number on it 1, 2, 3,
4, 5, 6, 7, 8, 9, 10. Let S set of all ordered
pairs of distinct chips, and C subset of S
with the first red and the second blue.By the
multiplication rule (drawn in order, w/out
replacement)
S 10 ? 9 90 and C 3 ? 7 21 so P(C)
C / S 21/90 7/30.
S 10 ? 9 90 and C _ ? _
S __ ? __
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.
Example. A bowl contains 7 blue chips and 3 red
chips. Two chips are drawn at random, in order
and without replacement. Find the probability
that the first is red and the second is blue. 2.
Use conditional probability. This method follows
from a rearrangement of the equation defining
conditional probability. If we multiply both
sides by P(B), we obtain the Multiplicative Law
of Probability P(A?B) P(B) P(AB). Similarly,
if we interchange the roles of A and B, we
obtain P(A?B) P(A) P(BA).
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.
P(A?B) P(A) P(BA)
Example. A bowl contains 7 blue chips and 3 red
chips. Two chips are drawn at random, in order
and without replacement. Find the probability
that the first is red and the second is blue.
Let S set of all ordered pairs of distinct
chips as before, A be the event that the first
chip drawn is red, and B be the event that the
second chip drawn is blue. Then the desired
probability is P(A?B) P(A) P(BA).
P(A) 3/10 since 3 of the 10 chips in the
bowl are red P(BA)
P(BA) 7/9 since 7 of the 9 chips left in
the bowl are blue so that P(A?B)
so that P(A?B) P(A) P(BA) (3/10)(7/9)
21/90 7/30
P(A)