Probability concepts by Sibanand Pattnaik - Part 3

• CONDITIONAL PROBABILITY

Untill now in probability, we have discussed the methods of finding the probability of events. If we have two events from the same sample space, does the information about the occurrence of one of the events affect the probability of the other event? Let us try to answer this question by taking up a random experiment in which the outcomes are equally likely to occur.

Consider the experiment of tossing three fair coins. The sample space of the experiment is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Since the coins are fair, we can assign the probability 1/8 to each sample point. Let E be the event ‘at least two heads appear’ and F be the event ‘first coin shows tail’.

Then
E = {HHH, HHT, HTH, THH}
and F = {THH, THT, TTH, TTT}

Therefore P(E) = P ({HHH}) + P ({HHT}) + P ({HTH}) + P ({THH})
= 1/8 + 1/8 + 1/8 + 1/8 = 1/2 (why??)
and P(F) = P ({THH}) + P ({THT}) + P ({TTH}) + P ({TTT})
= 1/8 + 1/8 + 1/8 + 1/8 = 1/2
Also E ∩F = {THH}
with P(E ∩F) = P({THH}) = 1/8

Now, suppose we are given that the first coin shows tail, i.e. F occurs, then what is the probability of occurrence of E? With the information of occurrence of F, we are sure that the cases in which first coin does not result into a tail should not be considered while finding the probability of E. This information reduces our sample space from the set S to its subset F for the event E. In other words, the additional information really amounts to telling us that the situation may be considered as being that of a new random experiment for which the sample space consists of all those outcomes only which are favourable to the occurrence of the event F.

Now, the sample point of F which is favourable to event E is THH.
Thus, Probability of E considering F as the sample space = 1/ 4 ,
or Probability of E given that the event F has occurred = 1/ 4

This probability of the event E is called the conditional probability of E given that F has already occurred, and is denoted by P (E|F).

Thus P(E|F) = 1/4

Note that the elements of F which favour the event E are the common elements of E and F, i.e. the sample points of E ∩F.
Thus, we can also write the conditional probability of E given that F has occurred as
P(E|F) = Number of elementary events favourable to E∩ F/ Number of elementary events which are favourable to F

DEFINITION CONDITIONAL PROBABILITY

Definition 1
If E and F are two events associated with the same sample space of a random experiment, the conditional probability of the event E given that F has occurred, i.e. P (E|F) is given by
P(E|F) = P(E ∩ F) / P(F) provided P(F) ≠0

Properties of conditional probability

Let E and F be events of a sample space S of an experiment, then we have

Property 1 :
P(S|F) = P(F|F) = 1
We know that
P(S|F) = P(S ∩F)/ P(F) = P(F)/ P(F) = 1
Also P(F|F) = P(F∩ F)/ P(F) = 1
Thus P(S|F) = P(F|F) = 1

Property 2 :
If A and B are any two events of a sample space S and F is an event of S such that P(F) ≠0, then
P((A ∪B)|F) = P(A|F) + P(B|F) – P ((A ∩B)|F)
In particular, if A and B are disjoint events, then
P((A∪B)|F) = P(A|F) + P(B|F)

Property 3:
P(E′|F) = 1 − P (E|F)

If P(A) = 7/ 13 , P (B) = 9/ 13 and P (A ∩ B ) = 4/ 13 , evaluate P (A|B)

P(A|B)= P(A ∩B) / P(B) = (4/13)/ (9/13)

A family has two children. What is the probability that both the children are boys given that at least one of them is a boy ?

Let b stand for boy and g for girl. The sample space of the experiment is
S = {(b, b), (g, b), (b, g), (g, g)}
Let E and F denote the following events :
E : ‘both the children are boys’
F : ‘at least one of the child is a boy’
Then E = {(b,b)} and F = {(b,b), (g,b), (b,g)}
Now E ∩F = {(b,b)}
Thus P(F) = 3/4
and P (E ∩F) = 1/ 4
Therefore P(E|F) = P(E ∩ F) / P(F) = (1/4)/(3/4) = 1/3

Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Let A be the event ‘the number on the card drawn is even’ and B be the event ‘the number on the card drawn is greater than 3’. We have to find P(A|B).
Now, the sample space of the experiment is S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Then A = {2, 4, 6, 8, 10}, B = {4, 5, 6, 7, 8, 9, 10}
and A ∩B = {4, 6, 8, 10}
Also P(A) = 5 /10 , P(B) = 7/10 and P(A ∩ B ) = 4/10
Then p(A|B) = P(A ∩ B ) / P(B) = (4/10)/ (7/10)

In a school, there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in Class XII given that the chosen student is a girl?

Ans - 1/10

A die is thrown three times. Events A and B are defined as below:
A : 4 on the third throw
B : 6 on the first and 5 on the second throw
Find the probability of A given that B has already occurred.

The sample space has 216 outcomes.
Now A =
(1,1,4) (1,2,4) ... (1,6,4) (2,1,4) (2,2,4) ... (2,6,4)
(3,1,4) (3,2,4) ... (3,6,4) (4,1,4) (4,2,4) ...(4,6,4)
(5,1,4) (5,2,4) ... (5,6,4) (6,1,4) (6,2,4) ...(6,6,4)
B = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)}
and A ∩B = {(6,5,4)}.
Now P(B) = 6/ 216
And P(A ∩ B ) = 1/216
Then p(A|B) = P(A ∩ B ) / P(B) = (1/216)/(6/216) = 1/6

A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Let E be the event that ‘number 4 appears at least once’ and F be the event that ‘the sum of the numbers appearing is 6’.
Then, E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)}
and F = {(1,5), (2,4), (3,3), (4,2), (5,1)}
We have P(E) = 11/ 36
and P (F) = 5/ 36
Also E∩F = {(2,4), (4,2)}
Therefore P(E∩F) = 2/ 36
Hence, the required probability
P(E|F) = P(E ∩ F) / P(F) = (2/36)/ (5/36) = 2/ 5

For the conditional probability, we have considered the elementary events of the experiment to be equally likely and the corresponding definition of the probability of an event was used. However, the same definition can also be used in the general case where the elementary events of the sample space are not equally
likely, the probabilities P (E∩F) and P (F) being calculated accordingly. Let us take up the following example.

Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that ‘the die shows a number greater than 4’ given that ‘there is at least one tail’.

The sample space of the experiment may be described as
S = {(H,H), (H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
where (H, H) denotes that both the tosses result into
head and (T, i) denote the first toss result into a tail and
the number i appeared on the die for i = 1,2,3,4,5,6.
Thus, the probabilities assigned to the 8 elementary
Events (H, H), (H, T), (T, 1), (T, 2), (T, 3) (T, 4), (T, 5), (T, 6)
Are 1/4, 1/4, 1/12, 1/12, 1/12, 1/12, 1/12, 1/12,
respectively.
Let F be the event that ‘there is at least one tail’ and E be the event ‘the die shows
a number greater than 4’. Then
F = {(H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
E = {(T,5), (T,6)} and E ∩F = {(T,5), (T,6)}
Now P(F) = P({(H,T)}) + P ({(T,1)}) + P ({(T,2)}) + P ({(T,3)}) + P ({(T,4)}) + P({(T,5)}) + P({(T,6)})
= 1/4 + 1/12 + 1/12 + 1/12 +1/12+ 1/12+1/12 = 3/4
and P(E ∩F) = P ({(T,5)}) + P ({(T,6)}) = 1/12 +1/12 = 1/6
Hence, the required probability
P(E|F) = P(E ∩ F) / P(F) = (1/6)/ (3/4) = 2/9.

The outcomes of the experiment can be represented in following diagrammatic manner called the ‘tree diagram’.

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