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Conditional Probability
P(A/B )=
𝑃(𝐴∩𝐵)
A
A∩𝐵
B
𝑃(𝐵)
- Knowing that event B has occurred, S is reduced to B
Figure1
- Using relative frequency repeated n-times.
𝑛𝐴∩𝐵/𝑛
𝑃(𝐴∩𝐵)
𝑛𝐵/𝑛
𝑃(𝐵)
= P(A/B)
Note
-
𝑃(𝐴 ∩ 𝐵) = P(A/B)P(B)= P(B/A)P(A)
Similarly,
-
P(A∩B ∩ 𝐶) = P(A/B)
= P(A/B∩ 𝐶)𝑃(𝐵 ∩ 𝐶)P(C)
Example:
In a binary communication system, suppose that P(A0)=1-P, P(A1)=P and P(error)=ε. Fined
𝑃(𝐴𝑖 ∩ 𝐵𝑗 )
where, i,j=0,1
Sol.
P(A0∩B0)=P(B0/A0)P(A0 ) = (1-ε)(1-P)
P(error)= ε =P(B0/A1)=P(B1/A0)
P(A0∩B1)=P(B1/A0)P(A0)= ε(1-P)
P(A1∩B0)=P(B0/A1)P(A1)= ε * P
P(A1∩B1)=P(B1/A1)P(A1)= (1-ε)P
Total Probability theorem
Let S=B1 ∩ B2 …… ∩ Bn (mutually exclusive events)
B1, B2, ………..,Bn are called partition of S
Any event A=A∩S = A∩(B1∪B2∪B3…….∪Bn)
= (A∩B1) ∪ (A∩B2) ∪ (A∩B3)…… ∪ (A∩Bn)
P(A)=P(A/B1)P(B1)+……………….+P(A/Bn)P(B)
This is useful when an experiment can be viewed as sequences of two sub-experiments as
shown in figure 2.
……….
Bn-1
B1
A
B2
Bn
B3
…
Figure2
Example:
A factory produces a mix of good and bad chips. The lifetime of good chips follow exponential
law with rate α and bad chips with rate 1000 α.
Suppose (1-P) of hips are good, find the probability that a randomly selected chip is still
functioning after t?
Sol.
G=” Chip is good”
B=” Chip is bad”
C=” Chip is functioning after time t ”
P(C/G) = 𝑒 −𝛼𝑡 , P(C/B) = 𝑒 −1000𝛼𝑡
Find P(C)
P(C) =P(C∪G) + P(C∩B)
=P(C/G)P(G)+P(C/B)P(B)
= 𝑒 −𝛼𝑡 (1-P) + 𝑒 −1000𝛼𝑡 P
Bayes’ Rule
Example:
Let S=B1∪B2∪ ………………….∪Bn (Partition)
Suppose that event A occurs, what is the probability of Bj?
Sol.
P(𝐵𝑗 /A) =
𝑃(𝐴∩𝐵𝑗 )
𝑃(𝐴)
𝐴
)𝑃(𝐵𝑗 )
𝐵𝑘
𝐴
∑𝑛
𝑘=1 𝑃(𝐵 )𝑃(𝐵𝑘 )
𝑘
𝑃(
=
Bayes’ rule is useful in finding the “a posteriori”
(P(Bj/A) in terms of the “ a priori” (P(Bj) before experiment is performed) and occurrence of A
Example:
In binary communication system, find which input is more probable given that the receive has
output 1. Assume P(A0) = P(A1) and P(error)= ε.
Sol.
P(A0/B1) =
=𝜀
2
𝜀∗12
(1−𝜀)
+
2
𝐵
𝑃( 1)𝑃(𝐴0 )
𝐴0
𝑃(𝐵1 )
=
𝐵
𝑃( 1)𝑃(𝐴0 )
𝐴0
𝐵1
𝐵
𝑃( )𝑃(𝐴0 )+𝑃( 1)𝑃(𝐴1 )
𝐴0
𝐴1
=ε
P(A1/B1) =
𝐵
𝑃( 1)𝑃(𝐴1 )
𝐴1
𝑃(𝐵 )
=
(1−)𝜀∗12
𝜀 (1−𝜀)
+
2
2
If ε<12, then A1 is more probable.
= 1-ε
Independence of events
Events A and B are independent if
P(A/B)= P(A)P(B) or P(A/B) = P(A)
Example: Two number x and y are selected at random between 0 and 1. Let event A={x> 0.5},
B={y<0.5}
C={x>y}
Are A and B independent?
Are A and C independent?
Sol.
P(A)=0.5=P(B)=P(C)
P(C∩B) = 14
P(A∩B) = 14
P(A∩C) = 38
P(A∩B) = 14 = P(A)P(B)
then A & B Independent.
P(A∩C) = 38 ≠P(A)P(B)
then A & C dependent.
Notes
(1): If P(A) > 0 and A and B are mutually exclusive or disjoint, then A and B cannot be
independent.
(2) If A and B are independent and mutually exclusives, then P(A)=0 or P(B)=0.
(3) A, B and C are independent, if:
P(A∩B)= P(A)P(B)
P(A∩C)= P(A)P(C)
P(B∩C)= P(B)P(C)
P(A∩B∩ 𝐶)= P(A)P(B)P(C)
Pairwise Independent
(4) Independent is often assumed if events have no obvious physical relation.
Sequential experiments
-
Many experiments can be decomposed of sequences of simpler sub experiments, which
may or may not be independent.
Sequences of independent experiments A1, A2, ………, Ak be events associated with the
outcomes of k independent sub-experiments S1, S2, …….., Sk then
P(A1∩A2… … … ∩ 𝐴k)= P(A1)P(A2)……P(Ak)
Example:
Suppose 10 numbers are selected at random between [0 1]. Find the probability that the first 5
numbers < 14 and the last 5 ≥ 12
Sol.
1 3
1 3
Probability (4) (2) = 3.05 *10−3
The binomial law
A Bernoulli trial involves performing as experiment and noting whether event A occurs
“success” or not “failure”.
Theorem: Let k be the number of success in n independent Bernoulli trials, then the probability
of k successes:
Pn(k)= (𝑛𝑘)𝑝𝑘 (1 − 𝑝)𝑛−𝑘−3
where k = 0, …………n
Example:
let k be the number of active speakers in a group of 8 independent speakers.
Suppose a speaker is active with probability = 13
Find probability that the number of active speakers is greater than 6.
Sol.
Let k=number of active speakers.
P(k>6) = P(k=7) + P(k=8)
1 7 2 1
1 8 2 0
= (87) (3) (3) + (88) (3) (3) = 0.00259
The multinomial probability law
Let the experiment is repeated n times and each Bj occurred kj times. Then, probability of the
vector (k1, k2, …………kn)
P(k1, k2, …………kn) =
𝑛!
𝑘1!𝑘2!……..𝑘𝑚!
𝑝1𝑘1 𝑝2𝑘2 ………. 𝑝𝑚𝑘𝑚
Example:
Pick 10 phone numbers at random and note the last digit. What is the probability that you
obtain 0 to 9 (with ordering)
Probability =
10!
1!1!……..1!
(0.1)10 = 3.6 * 10−4
The Geometric probability law
-
Here, the outcome is the number of independent Bernoulli trials until the occurrence of
the first success. S= {1, 2, …………∞}
The probability P(m) that m trials are required means that the first m-1 result in failures
and the m’th trials is success. Let Ai = “success in i’th trials”
P(m) = P(A1, A2, ………..Am-1, An) = (1 − 𝑝)𝑚−1 p
-
where n=1,2, ……….
The probability that more than k trials are required before a success:
∞
∞
P[{m>K}] = p ∑𝑚=𝑘+1 𝑞 𝑚−1 = p 𝑞 k ∑𝑗=0 𝑞 𝑗
= p 𝑞k
= 𝑞k
1
1−𝑞
Example:
Computer A sends a message to computer B. If the probability of error is (0.1). if error
occurs, B request A resend the message. What is the probability that a message is
transmitted more than twice?
Sol.
Each transmission is Bernoulli trial with P(success) = 0.9 and P(failure) =0.1
P(m>2) = (1 − p)2 (1 − 0.9)2 = (0.1)2 = 0.01