`color{red}♦` Conditional Probability

`color{red}♦` Properties of conditional probability

`color{red}♦` Properties of conditional probability

`=>` Consider the experiment of tossing three fair coins. The sample space of the experiment is

`color {orange} {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`

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) = ( text (Number of elementary events favourable to E ∩ F ) )/( text(Number of elementary events which are favourable to F) )`

`color{blue}{= ( n ( (E∩ F) ) )/(n (F) )}`

`=>` Dividing the numerator and the denominator by total number of elementary events of the sample space, we see that `P(E/F)` can also be written as

`color{red}{P(E|F) = ( ( n(E ∩ F) )/(n(S)) )/( (n(F))/(n(S))) = ( P(E∩ F))/( P(F) )}` ......(1)

Note that (1) is valid only when `P(F) ≠ 0` i.e., `F ≠ φ`

Thus, we can define the conditional probability as follows :

`color {blue} "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

`color {orange} {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`

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) = ( text (Number of elementary events favourable to E ∩ F ) )/( text(Number of elementary events which are favourable to F) )`

`color{blue}{= ( n ( (E∩ F) ) )/(n (F) )}`

`=>` Dividing the numerator and the denominator by total number of elementary events of the sample space, we see that `P(E/F)` can also be written as

`color{red}{P(E|F) = ( ( n(E ∩ F) )/(n(S)) )/( (n(F))/(n(S))) = ( P(E∩ F))/( P(F) )}` ......(1)

Note that (1) is valid only when `P(F) ≠ 0` i.e., `F ≠ φ`

Thus, we can define the conditional probability as follows :

`color {blue} "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

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

`●color {blue} "Property 1"` ` color {orange} {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)) = (P(F))/(P(F) ) =1`

Thus `P(S|F) = P(F|F) = 1`

`●color {blue} "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

`color {orange} {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)`

We have

`P((A∪B)|F) = (P [A (A ∪ B) ∩ F ] )/(P(F) )`

`= (P [ (A ∩ F ) ∪ (B ∩ F) ] )/( P (F) )`

(by distributive law of union of sets over intersection)

`= ( P(A ∩ F)+P(B ∩ F)–P(A ∩ B ∩F )/(P(F))`

`= ( p (A ∩ F)/(P(F)) + ( P ( B ∩ F ) )/(P (F) ) - ( P [ (A ∩ B) ∩ F ] )/( P (F) )`

`= P(A|F) + P(B|F) – P ((A∩B)|F) `

When A and B are disjoint events, then

`P((A ∩ B)|F) = 0`

`⇒ P((A ∪ B)|F) = P(A|F) + P(B|F)`

`●color{blue} "Property 3"` `color {orange} { P(E′|F) = 1 − P (E|F)}`

From Property 1, we know that P (S|F) = 1

`⇒ P(E ∪ E′|F) = 1` since `S = E ∪ E′`

`⇒ P(E|F) + P (E′|F) = 1` since E and E′ are disjoint events

Thus,` P(E′|F) = 1 − P(E|F)`

`●color {blue} "Property 1"` ` color {orange} {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)) = (P(F))/(P(F) ) =1`

Thus `P(S|F) = P(F|F) = 1`

`●color {blue} "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

`color {orange} {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)`

We have

`P((A∪B)|F) = (P [A (A ∪ B) ∩ F ] )/(P(F) )`

`= (P [ (A ∩ F ) ∪ (B ∩ F) ] )/( P (F) )`

(by distributive law of union of sets over intersection)

`= ( P(A ∩ F)+P(B ∩ F)–P(A ∩ B ∩F )/(P(F))`

`= ( p (A ∩ F)/(P(F)) + ( P ( B ∩ F ) )/(P (F) ) - ( P [ (A ∩ B) ∩ F ] )/( P (F) )`

`= P(A|F) + P(B|F) – P ((A∩B)|F) `

When A and B are disjoint events, then

`P((A ∩ B)|F) = 0`

`⇒ P((A ∪ B)|F) = P(A|F) + P(B|F)`

`●color{blue} "Property 3"` `color {orange} { P(E′|F) = 1 − P (E|F)}`

From Property 1, we know that P (S|F) = 1

`⇒ P(E ∪ E′|F) = 1` since `S = E ∪ E′`

`⇒ P(E|F) + P (E′|F) = 1` since E and E′ are disjoint events

Thus,` P(E′|F) = 1 − P(E|F)`

Q 3127291181

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

Class 12 Chapter 13 Example 1

Class 12 Chapter 13 Example 1

We have` P (A | B ) = ( P(A B) )/( P(B) ) = (4/13)/(9/13) = 4/9`

Q 3147291183

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 ?

Class 12 Chapter 13 Example 2

boys given that at least one of them is a boy ?

Class 12 Chapter 13 Example 2

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`

Q 3177291186

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?

Class 12 Chapter 13 Example 3

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?

Class 12 Chapter 13 Example 3

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) = 4/7`

Q 3117291189

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?

Class 12 Chapter 13 Example 4

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?

Class 12 Chapter 13 Example 4

Let E denote the event that a student chosen randomly studies in Class XII

and F be the event that the randomly chosen student is a girl. We have to find P (E|F).

Now P(F) = 430/1000 = 0.43 and P(E∩ F) = 43/1000 = 0.043 (why ? )

Then P(E|F) = (P(E∩ F) )/(P(F)) = (0.043)/( 0.43) = 0.1`

Q 3127391281

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.

Class 12 Chapter 13 Example 5

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.

Class 12 Chapter 13 Example 5

The sample space has 216 outcomes.

Now `A = { tt ( ((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`

Q 3157391284

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?

Class 12 Chapter 13 Example 6

to be 6. What is the conditional probability that the number 4 has appeared at least

once?

Class 12 Chapter 13 Example 6

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 discussed above, 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.

Q 3117491380

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’.

Class 12 Chapter 13 Example 7

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’.

Class 12 Chapter 13 Example 7

The outcomes of the experiment can be

represented in following diagrammatic manner called

the ‘tree diagram’.

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 which is clear from the Fig 13.2.

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 + 1/2 + 1/12+ 1/12+1/2+1/12+1/12 = 3/4`

and P(E ∩ F) = P ({(T,5)}) + P ({(T,6)}) = 1/12 +1/12 =1/6

Hence P(E|F) = (P(E ∩ F) )/( P(F)) = (1/6)/(3/4) = 2/9`