5. Conditional probability
Many times we know additional information that affects the calculation of
a probability:
• What is the probability that a person voted for Obama if you know he is
a registered Republican?
• What is that probability that the a person is guilty of murder if you know
the crime was committed by a left-handed person? If the person is right
handed, the probability is 0. If the person is left handed, is the probability
is 1?
The probability that A occurs given that B occurs is called the conditional
probability of A given B and is
written P(A|B) as
P(𝐴∩𝐵)
P(A|B) =
; for P(B) > 0:
P(𝐵)
Note that we can only condition on events with positive probability.
Under the classical interpretation of probability, we can see that if we are
told that B has occurred, then all outcomes in B are equally likely, and all
outcomes not in B have zero probability
Example: A fair coin is tossed two times. What is the probability that the
second coin is a head if you know that at least one head appears.
The four outcomes are:
HH ,HT,TH
TT
We have checked those outcomes which satisfy the condition that at least
one head occurs.
Note that the fourth outcome (TT) in the table is not checked. Out of the
three events that satisfy this condition (HH, HT, TH) exactly two have
heads for the second toss
P(second toss is a head given at least one head) = 2/ 3
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Example:- Lawyers and Liars Cont’d: suppose There are 100 people at a party,
Forty are liars, Twenty-five are lawyers, 15 of the lawyers are liars. If a person is
drawn at random at the party
(say to win a door prize), what is the probability that he or she is a lawyer?
Sol:_
Since there are 25 lawyers out of 100 people,
the probability is P(lawyer) =25/100=1/4
Example:- Lawyers and Liars Cont’d: suppose There are 100 people at a party,
Forty are liars, Twenty-five are lawyers, 15 of the lawyers are liars. If a person is
drawn at random at the party. what is the probability that he or she is a lawyer if
you know that he or she is a liar?
We are restricted to the subset of forty liars.
Since we are told that 15 of these 40 people are also lawyers,
P(lawyer | liar) =15/40=3/8.
Example. A fair coin is tossed two times. What is the probability that the second
coin is a head if you know that at least one head appears.
The four outcomes are:
(1) H H (2) H T
(3)T H (4) T T
Let A be the event “at least one head”
and B be the event “the second
toss is a head”
By direct count P(B) = 2/4, P(A) = ¾, P(A ∩ B) = 2/4
So by the Conditional Probability Rule
P(B | A) = P(A ∩ B)/P(A) = 2/4/3/4 = 2/3.
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Some proprieties of conditional probability
P(Ω |B) =P(Ω ∩ B)/P(B)=P(B)/P(B)= 1,
since we have P(B |B) =P(B)/P(B) = 1,
To verify the additivity axiom, we write for any two disjoint events A1 and A2,
P(A1 ∪ A2 |B) =P((A1 ∪ A2) ∩ B)/P(B)
=P ((A1 ∩ B) ∪ (A2 ∩ B))/P (B)
=P (A1 ∩ B) + P(A2 ∩ B)/P(B)
=P (A1 ∩ B)/P(B)+P(A2 ∩ B)/P(B)
= P(A1 |B) + P(A2 |B),
Where for the second equality, we used the fact that A1 ∩ B and A2 ∩ B are
disjoint sets, Since conditional probabilities constitute a legitimate probability law,
all general properties of probability laws remain valid.
For example,
a fact such as P(A ∪ C) ≤ P(A) + P(C) translates to the new fact P(A ∪ C |B) ≤
P(A|B) + P(C |B).
Example:- We toss a fair coin three successive times. We wish to find the
conditional probability P(A|B) when A and B are the events
A = {more heads than tails come up}, B= {1st toss is a head}.
The sample space consists of eight sequences, Ω = {HHH, HHT, HTH, HTT,
THH, THT, TTH, TTT}, which we assume to be equally likely.
The event B consists of the four elements HHH, HHT, HTH, HTT, so its
probability is P(B) =4/8.
The event A ∩ B consists of the three elements outcomes HHH, HHT, HTH, so its
probability is P(A ∩ B) =3/8
Thus, the conditional probability P (A|B) is P (A|B) =P (A ∩ B)/P (B)
=3/8/4/8=3/4
Because all possible outcomes are equally likely here, we can also compute
P (A|B)
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