Discrete Probability
Conditional Probability
R. Inkulu
http://www.iitg.ac.in/rinkulu/
(Conditional Probability)
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Definition
Let H be an event with positive probability. For an arbitrary event A, the
conditional probability of A on the hypothesis H (or for given H) is denoted
1
with p(A|H), and is equal to p(A∩H)
p(H) .
subsample space H becomes our new sample space with probabilities proportional to the original
ones; p(H) is necessary in order to reduce the total probability of the new sample space to unity
An example: Consider families with exactly two children. Given that a family
has a boy, the probability that both children are boys is
1
The conditional probability p(B|A) is called a posteriori if event B precedes event A in
time; otherwise, it is called a priori.
(Conditional Probability)
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Definition
Let H be an event with positive probability. For an arbitrary event A, the
conditional probability of A on the hypothesis H (or for given H) is denoted
1
with p(A|H), and is equal to p(A∩H)
p(H) .
subsample space H becomes our new sample space with probabilities proportional to the original
ones; p(H) is necessary in order to reduce the total probability of the new sample space to unity
An example: Consider families with exactly two children. Given that a family
has a boy, the probability that both children are boys is 13 .
1
The conditional probability p(B|A) is called a posteriori if event B precedes event A in
time; otherwise, it is called a priori.
(Conditional Probability)
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Few consequences
• product rule of conditional probability:
p(A ∩ H) = p(A|H)p(H)
generalizing the same yields,
p(E1 ∩E2 ∩. . .∩En ) = p(E1 |E2 ∩E3 ∩. . .∩En )p(E2 |E3 ∩. . .∩En ) . . . p(En )
• p(A ∪ B|H) = p(A|H) + p(B|H) − p((A ∩ B)|H)
• law of total probability:
let H1 , . . . , Hn be a set of mutually exclusive events that partition the
sample space, then for any arbitrary event A,
P
p(A) = nj=1 p(A|Hj )p(Hj )
(Conditional Probability)
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Bayes’ rule
• let H1 , . . . , Hn be a set of mutually exclusive events that partition the
sample space, then for any arbitrary event A,
p(Hk |A) =
Pp(A|Hk )p(Hk ) .
j p(A|Hj )p(Hj )
(Conditional Probability)
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Kinds of random sampling
• When the ball selected is not returned to the bin before the next ball is
selected then this sampling is termed as sampling without replacement.
Typically the probability of selecting a ball from the bin changes from
one iteration to another.
• Otherwise, if the ball selected is returned to the bin before the next ball is
selected then this sampling is termed as sampling with replacement.
2
2
when compared with sampling without replacement, sampling with replacement is often
simpler to code and the effect on the probability of making an error is almost negligible, hence
making it a desirable alternative
(Conditional Probability)
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Outline
1
Examples
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• Four balls are successively placed into four cells. Given that the first two
balls are in different cells, the probability that one cell contains exactly
three balls is
(Conditional Probability)
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• Four balls are successively placed into four cells. Given that the first two
balls are in different cells, the probability that one cell contains exactly
three balls is 422 .
(Conditional Probability)
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• The probability of a family having exactly k children is pk (where
P
pk = 1). If it is known that a family has boys but no girls, the
(conditional) probability that it has one only one child is
(Conditional Probability)
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• The probability of a family having exactly k children is pk (where
P
pk = 1). If it is known that a family has boys but no girls, the
−1
(conditional) probability that it has one only one child is Pp1 2pj 2−j .
j
(Conditional Probability)
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Polya’s urn scheme
• A bin contains b black and r red balls. A ball is drawn at random. It is
replaced and, moreover, c > 0 balls of the color drawn are added to the
bin. A new random sampling is made from the bin, and this procedure is
repeated. The probability that of n = n1 + n2 drawings, the first n1 ones
result in black balls and the remaining n2 ones in red balls is
(Conditional Probability)
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Polya’s urn scheme
• A bin contains b black and r red balls. A ball is drawn at random. It is
replaced and, moreover, c > 0 balls of the color drawn are added to the
bin. A new random sampling is made from the bin, and this procedure is
repeated. The probability that of n = n1 + n2 drawings, the first n1 ones
result in black balls and the remaining n2 ones in red balls is
b(b+c)(b+2c)...(b+(n1 −1)cr(r+c)...(r+(n2 −1)c)
(b+r)(b+r+c)(b+r+2c)...(b+r+(n−1)c)
(Conditional Probability)
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Polya’s urn scheme
• A bin contains b black and r red balls. A ball is drawn at random. It is
replaced and, moreover, c > 0 balls of the color drawn are added to the
bin. A new random sampling is made from the bin, and this procedure is
repeated. The probability that of n = n1 + n2 drawings, the first n1 ones
result in black balls and the remaining n2 ones in red balls is
b(b+c)(b+2c)...(b+(n1 −1)cr(r+c)...(r+(n2 −1)c)
(b+r)(b+r+c)(b+r+2c)...(b+r+(n−1)c)
in fact, any other orderings of n1 black and n2 red balls has the same
probability
(Conditional Probability)
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• Let a bin I contains r1 red and b1 black balls, and an bin II contains r2
red and b2 black balls. A die is thrown; if ace appears, choose bin I,
otherwise bin II. Thereafter random drawings with replacement are done
from the chosen bin. Given that the first drawing resulted in red, the
(conditional) probability of a sequence red, red is
(Conditional Probability)
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• Let a bin I contains r1 red and b1 black balls, and an bin II contains r2
red and b2 black balls. A die is thrown; if ace appears, choose bin I,
otherwise bin II. Thereafter random drawings with replacement are done
from the chosen bin. Given that the first drawing resulted in red, the
(conditional) probability of a sequence red, red is
r
r2
1
( 1 )2 + 56 ( r +b
)2
6 r1 +b1
2
2
r
r
1
2 )
( 1 )+ 56 ( r +b
6 r1 +b1
2
2
(Conditional Probability)
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Law of succession of Laplace
• Imagine a collection of n + 1 bins, each containing a total of n red and
white balls; the bin number i contains i red and n − i white balls for
i = 0, 1, 2, . . . , n. An bin is chosen at random and k random drawings are
made from it, the ball drawn being replaced each time. Given that all k
balls turn out to be red, the (conditional) probability that the next
drawing will also yield a red ball is
(Conditional Probability)
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Law of succession of Laplace
• Imagine a collection of n + 1 bins, each containing a total of n red and
white balls; the bin number i contains i red and n − i white balls for
i = 0, 1, 2, . . . , n. An bin is chosen at random and k random drawings are
made from it, the ball drawn being replaced each time. Given that all k
balls turn out to be red, the (conditional) probability that the next
drawing will also yield a red ball is
1n+1 +2n+1 +...+nk+1
nk+1 (n+1)
1n +2n +...+nk
nk (n+1)
when n is lage, the above is approximately,
(Conditional Probability)
n+1
n+2
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• In answering a question on a multiple-choice test, a student either knows
the answer or guesses. Let p be the probability that the student knows the
answer and (1 − p) the probability that the student guesses. Assume that
a student who guesses at the answer will be correct with the probability
1
m , where m is the number of multiple-choice alternatives. The
conditional probability that a student knew the answer to a question,
given that he or she answered it correctly is
(Conditional Probability)
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• In answering a question on a multiple-choice test, a student either knows
the answer or guesses. Let p be the probability that the student knows the
answer and (1 − p) the probability that the student guesses. Assume that
a student who guesses at the answer will be correct with the probability
1
m , where m is the number of multiple-choice alternatives. The
conditional probability that a student knew the answer to a question,
mp
given that he or she answered it correctly is 1+(m−1)p
.
(Conditional Probability)
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• An insurance company believes that people can be divided into two
classes: those who are accident prone and those who are not. An
accident-prone person will have an accident at some time with in a fixed
one-year period with probability p whereas the probability decreases to q
for a non-accident-prone person. Assume that r is the probability of
being accident prone. Supposing that a new policyholder has an accident
within an year of purchasing a policy, the probability that he or she is
accident prone is
(Conditional Probability)
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• An insurance company believes that people can be divided into two
classes: those who are accident prone and those who are not. An
accident-prone person will have an accident at some time with in a fixed
one-year period with probability p whereas the probability decreases to q
for a non-accident-prone person. Assume that r is the probability of
being accident prone. Supposing that a new policyholder has an accident
within an year of purchasing a policy, the probability that he or she is
rp
accident prone is rp+(1−r)q
(Conditional Probability)
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