Conditional Probability
The probability of
occurrence of an event B when it is known that some event A has occurred is
called a condition probability and is denoted by P(B/A). The symbol P(B/A) is
usually read '‘the probability that B occurs given that A occurs” or ”simply
probability of B, given A”.
Consider two events ‘A’ and
‘B’ of samplespace S. When it is known that event ‘A’ has occurred, it means
that sample space would reduce to the sample points representing event A. Now
for P(B/A) we must look for the sample points representing the simultaneous
occurrence of A and B i.e. sample points in AÇB.
&⇒ P(B/A) =
Thus P(B/A) = ,
where 0 < P(A) £ 1
Similarly, P(A/B)
= 0
< P(B) £ 1
Hence, P(A Ç B) =
Consider the event
‘B’ of getting a ‘4’ when a fair die is tossed. Now suppose that it is known
that toss of die resulted in a number greater than 3 (say event A). And we have
to obtain P(B/A) i.e. the probability of getting a ‘4’ given that a number greater
than 3
has occurred. .
Clearly A = {4, 5,
6}, B = {4}
&⇒ P(B/A) =
also P(AÇB) = and
P(A) =
&⇒ P(B/A) = =
Example.1
The probability that a married man
watches a certain T.V. show is 0.4 and the probability that a married woman
watches the show is 0.5. The probability that a man watches the show, given
that his wife does, is 0.7. Find .
(a) the probability that married couple watch the show
(b) the probability that a wife watches the show given that
her husband does.
(c) the probability that atleast one person of a married
couple will watch the show.
Solution:
Let
‘H’ be the event that married man watches the show and ‘W’ be the event that
married woman watch the show,
&⇒ P(H) = 0.4, P(W) =
0.5, P(H/W) = 0.7.
(a) P(HÇW) = P(W)×P(H/W) = 0.5 ´ 0.7 = 0.35.
(b) P(W/H) =
(c) P(HÈW) = P(H) + P(W)  P(HÇW)
= 0.4 + 0.5  0.35 = 0.55.
Example.2
Consider the sample space ‘S’ representing the adults in
a small town who have completed the requirements for a college degree. They
have been categorised according to sex and employment as under:.

Employed

Unemployed

Male

460

40

Female

140

260

A employed person is selected at random.
Find the probability that chosen one is male.
Solution
Let M be the event
that man is chosen and E be the event that chosen one is employed.
From the concept of
reduced sample space we immediately get
P(M/E) =
This can be done
also as P(E) =
P(EÇM) =
&⇒ P(M/E) =
Example.3
A bag contains 3 white balls and 2 black
balls, another contains 5 white and 3 black balls. If a bag is chosen at random
and a ball is drawn from it. What is the probability that it is white.
Solution:
The
probability that the first bag is chosen is 1/2 and the chance of drawing a
white ball from it is 3/5.
\ Chance of choosing the first bag and
drawing a white ball is 1/2.3/5.
Similarly, the chance of choosing the
second bag and drawing a white ball is 1/2.5/8.
\
The chance of randomly choosing a bag and drawing a white ball is = (Mutually
exclusive cases) = 49/80.
Independent Events:
Two events A and B
are said to be independent if occurrence or nonoccurrence of one does not
affect the occurrence or nonoccurrence of the other,
i.e. P(B/A) = P(B),
P(A) > 0 similarly P(A/B)
= P(A), P(B) > 0.
&⇒ P(B/A) =
&⇒ P(AÇB) = P(A)×P(B)
If the events are
not independent, they are said to be dependent.
Remarks:
q
If
P(A) = 0
&⇒ for event ‘B’, 0 £ P(AÇB) £
P(A).
&⇒
P(AÇB) = 0, thus P(AÇB) = P(A)×P(B) = 0. Hence an impossible event
would be independent of any other event.
q
Distinction
between independent and mutually exclusive events must be carefully made. If A
and B are two mutually exclusive and possible events of sample space ‘S’ then
P(A) > 0, P(B) > 0 and P(AÇB)
= 0 > P(A)×P(B) so that A and B can’t be
independent. Infact P(A/B) = 0 similarly P(B/A) = 0. Consequently, mutually
exclusive events are strongly dependent. .
q
Two
events A and B are independent if and only if A and B¢ are independent or A¢ and B are independent or A¢ and B¢ are independent. .
We have P(AÇB) = P(A)×P(B)
Now P(AÇB¢) = P(A)  P(AÇB)
= P(A)  P(A)×P(B)
= P(A) (1P(B))
= P(A)×P(B¢)
Thus A and B¢ are independent.
Similarly P(A¢ÇB) = P(B) 
P(AÇB)
= P(B)×P(A¢)
Finally P(A¢ Ç
B¢) = 1  P(AÈB)
= 1  P(A)  P(B) + P(AÇB)
= 1  P(A)  P(B) + P(A)×P(B)
= (1P(A))  P(B) (1P(A))
= (1P(A)) (1P(B))
= P(A¢)×P(B¢)
Thus A¢ and B¢
are also independent.
Example.1
An event A_{1} can happen with probability p_{1}
and event A_{2} can happen with probability p_{2}.
What is the probability that
(i) exactly one of them happens
(ii) at least one of them happens(Given A_{1} and A_{2}
are independent events).
Solution:
(i) The probability that A_{1}
happens is p_{1}
\
The probability that A_{1} fails is 1  p_{1}
Also the probability that A_{2}
happens is p_{2}
Now, the chance that A_{1}
happens and A_{2} fails is p_{1}(1  p_{2}) and the
chance that A_{1} fails and A_{2} happens is p_{2}(1p_{1})
\
The probability that one and only one of them happens is
p_{1}(1  p_{2}) + p_{2}(1
 p_{1}) = p_{1} + p_{2}  2p_{1}p_{2}
(ii) The probability that both of
them fail = (1  p_{1}) (1  p_{2}).
\
probability that atleast one happens=1 (1  p_{1}) (1  p_{2})=p_{1}+p_{2}
 p_{1} p_{2}
Mutual Independence and
Pairwise Independence:
Three events A, B, C are said to be mutually
independent if,
P(AÇB)
= P(A)×P(B), P(A Ç C) = P(A)×P(C), P(B Ç C) = P(B)×P(C) and P(AÇBÇC)
= P(A)×P(B)×P(C)
These events would be said to be pairwise
independent if,
P(AÇB)
= P(A)×P(B), P(BÇC) = P(B)×P(C) and P(AÇC) = P(A)×P(C).
Thus mutually independent events are
pairwise independent but the converse may not be true.
Example.1
A
lot contains 50 defective and 50 nondefective bulbs. Two bulbs are drawn at
random, one at a time with replacement. The events A, B, C are defined as
under;.
A
= {the first bulb is defective}
B
= {the second bulb is non defective}
C
= {the two bulbs are both defective or both are nondefective}.
Categorise the events A, B, C to be
pairwise independent or mutually independent. .
Solution
Let D_{i} (N_{i})
denote the occurrence of a defective (nondefective) bulb at the i^{th}
draw (i = 1, 2). Now samplespace ‘S’ is; .
S = {D_{1} D_{2},
D_{1} N_{2}, N_{1} D_{2}, N_{1} N_{2}}
&⇒ P(D_{i}) = (i
= 1, 2)
A = {D_{1} D_{2},
D_{1} N_{2}}, B = {D_{1} N_{2}, N_{1} N_{2}}
, C = {D_{1} D_{2}, N_{1} N_{2}}
&⇒ AÇB
= {D_{1} N_{2}}, AÇC
= {D_{1} D_{2}}, BÇC
= {N_{1} N_{2}} and A Ç
B Ç C = f
P(A) = P(D_{1}
D_{2}) + P(D_{1}N_{2}) =
P(B) = P(D_{1}
N_{2}) + P(N_{1} N_{2}) =
P(C) = P(D_{1}
D_{2}) + P(N_{1}N_{2}) =
P(AÇB) = P(D_{1} N_{2}) ==
P(A)×P(B)
P(BÇC) = P(N_{1} N_{2}) ==
P(B)×P(C)
P(AÇC) = P(D_{1} D_{2})
= =
P(A)×P(C)
And P(A Ç B Ç C) = 0 >
P(A) × P(B) × P(C)
&⇒ Events are not mutually independent
but are pairwise independent.
Example.2
A person draws a card from a pack of 52,
replaces it and shuffles it. He continues doing it until he draws a spade. What
is the chance that he has to make .
(i) atleast 3 trials
(ii) exactly 3 trials :
Solution
(i) For
atleast 3 trials, he has to fail at the first 2 attempts and then after that it
doesn't make a difference if he fails or wins at the 3rd or the subsequent
attempts.
Chance of
success at any attempt = 1/4
\ chance of failure = 3/4
\ chance of failing in first 2 attempt
=
(ii) For
exactly 3 attempts, he has to fail in the first two attempts and succeed in the
3rd attempt.
\ required probability =