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Conditional Probability P (A/B) denotes the probability of event A happening, given that event B has happened. It is the conditional probability of A, given B. While calculating P (A/B), we assume that event B has occurred. It implies that the outcomes favorable to B become the total outcomes and hence outcomes favorable to P (A/B) are outcomes common to A and B. P (A/B) = P (A given B) = (Total no. of favorable cases)/ (Total no. of cases) = n (A∩B)/n (B) = P (A∩B)/P (B). Thus P (A/B) = P (A∩B)/P (B). => (P (A∩B) = P (B). (P (A/B) = P (A). (P(B/A)). Note that if A and B are independent events then P (A∩B) = P (A) P (B) => P (A/B) = P (A∩B)/P (B) = P (A) P (B)/P (B) = P (A) which should be the case as occurrence of A does not depend upon B. Similarly P (B/A) = P (B). Sometimes it also happens so that the probability of one event is affected by occurrence of other events. Let a bag contains 3 red, 2 green and 4 yellow balls. A ball is drawn and found to be red. Consider the following cases: (i) If the red ball is replaced then find the probability that the next ball to be drawn is yellow. (ii) What happens when red ball will not be replaced? Let, R: red ball is first drawn Y: yellow ball is drawn next. In first case when the red ball is replaced. P(Y) = (Total number of yellow balls)/ (total number of balls) =4/9 In second case the red ball is not replaced, so, P(Y) = (Total number of yellow balls)/ (Total number of balls left) = 4/8 The probability in these two cases is different. In the first case when the first ball was red and it replaced it does not affect the probability of yellow ball in second draw but in second case when the ball was not replaced the probability of yellow ball is changed. It happened because in second case the sample space is changed. So from here we can draw a conclusion that when occurrence of any event changes the sample space, the probability of other event change. Hence the probability also changes. It means that, occurrence of any event change the sample space. Then Probability of other events changes, in other words probability of some event depends on the occurrence of other events. This is known as "Conditional Probability". Let A and B be the two events. If A has already happened (i.e. given) then probability of B can be found by using the formula P (B/A) = P (A∩B)/P (A) If A and B are independent events, then P (B/A) = P (B) i.e. P (A∩B) = P (A) P (B) or, P(A/B) = P(A) (i) n events A_{1}, A_{2}, A_{3}, ........... A_{n} are said to be pair-wise independent tiff P (A_{i}∩A_{j}) = P(A_{i}) P(A_{j}), where 1, I = 1, 2, 3, ........., n (ii) These n events are said to be mutually independent iff P(A_{1} ∩ A_{2} ∩......... ∩ A_{n}) = P(A_{1}) P(A_{2}) ....... P(A_{n}). Remark: If the set of n events related to a sample space are pair-wise independent, they must be mutually independent, but vice versa is not always true. To apply total probability formula we must check whether A_{1} and A_{2} fulfill the following three conditions or not. A_{1} ∩ A_{2} = Φ and A_{1} U A_{2} = S. A_{1 }∈ S and A_{2 }∈ S. Probability of an event which depends on more than one event Suppose we have n events A_{1}, A_{2}, ........., A_{n} related to a sample space such that: (i) They are mutually exclusive i.e. A_{i} ∩ A_{j} = Φ, I, j = 1, 2 .......... n, I ≠ j. (ii) They are exhaustive i.e. A_{1} U A_{2} U A_{3} ......... U A_{n} = S. (iii) They are proper subsets of sample space S i.e. A_{i} S, i = 1, 2, .......... n. Then probability of an event A which depends on the events A_{1}, A_{2}, ........., An can be calculated as P (A) = ∑^{n}_{(i=1)} P(A/A_{i}) P(A_{i}) = P (A/A_{1}) P (A_{1}) + P (A/A_{2}) P (A_{2}) +......+ P (A/A_{n}) P (A_{n}) This is a formula for total probability. Illustration: Take the case of previous illustration in which a bag contains 3 Red, 2 Green and 4 Yellow balls. A ball is drawn and if it is Red or Yellow it is replaced otherwise not. Find the probability that second ball drawn is a yellow ball. Solution: First of all, we should be very clear about the event of which probability is to be found and the events on which it depends. Let's A_{1}: First ball is either red or yellow A_{2}: First ball is green A: Second ball drawn is yellow We have to calculate P(A) which definitely depends on A_{1} and A_{2}. Therefore, P(A) = P(A/A_{1}) P(A_{1}) + P(A/A_{2}) P(A_{2}) P(A/A_{2}) = Probability of A when A_{1} is already happen. = 4/9 (i.e. probability of A given A_{1}) P(A_{1}) = 7/9 P(A/A_{2}) = Probability of A when A_{2} is already happened = (Probability of A given A_{2}) = 4/8 P(A_{2}) = 2/9 Hence, P(A) = 4/9×7/9+4/8×2/9 = 37/81 View this video for more on conditional probability Example: In a card game, suppose a player needs to draw two cards of the same suit in order to win. Of the 52 cards, there are 13 cards in each suit. Suppose first the player draws a heart. Now the player wishes to draw a second heart. Since one heart has already been chosen, there are now 12 hearts remaining in a deck of 51 cards. So the conditional probability P (Draw second heart |First card was also a heart) = 12/51. Illustration: Suppose an individual applying to a college knows that there are 80% chances of his application being accepted and also there is a condition that only 60% of all the accepted students will be provided the dormitory housing. Hence, the chance of the student being accepted and receiving dormitory housing is given by P (Accepted and Dormitory Housing) = P (Dormitory Housing | Accepted) P (Accepted) = (0.60)*(0.80) = 0.48. Illustration: A jar contains red and green toys. Two toys are drawn without replacement. The probability of selecting a red toy and then a green toy is 0.34, and the probability of selecting a red toy on the first draw is 0.47. What is the probability of selecting a green toy in the second draw, given that the first toy drawn was red? Solution: Clearly it is a question of conditional probability. We need to find the probability of drawing a green toy having known that the first toy drawn was red. P (green | red) = P (red and green) / P (red) = 0.34 / 0.47 = 0.72 = 72%. Illustration: At ABC Senior Secondary school, the probability that a student takes Science and English is 0.087. The probability that a student takes Science is 0.68. What is the probability that a student takes English given that the student is taking Science? Solution: We need to find the probability of English given Science i.e. P (English | Science) = P (Science and English) / P (Science) = 0.087 / 0.68 = 0.13 = 13%. You may consult the Sample Papers to get an idea about the types of questions asked. Remark: To calculate the probability of the intersection of more than two events, the conditional probabilities of all of the preceding events must be considered. In the case of three events, A, B, and C, the probability of the intersection P(A and B and C) = P(A)P(B|A)P(C|A and B). Example: Consider the college applicant who has determined that he has 0.80 probability of acceptance and that only 60% of the accepted students will receive dormitory housing. Of the accepted students who receive dormitory housing, 80% will have at least one roommate. The probability of being accepted and receiving dormitory housing and having no roommates is calculated by: P(Accepted and Dormitory Housing and No Roommates) = P(Accepted)P(Dormitory Housing|Accepted)P(No Roomates|Dormitory Housing and Accepted) = (0.80)*(0.60)*(0.20) = 0.096. The student has about a 10% chance of receiving a single room at the college. Illustration: If a dice is thrown, what is the probability of occurrence of a number greater than 1, if it is known that only odd numbers can come up. Solution: S = the sample space = {1, 2, 3, 4, 5, 6} A = the event of occurrence of an odd number = {1, 3, 5} B = the event of occurrence of a number greater than 1= {2, 3, 4, 5, 6} Here A∩B = {3, 5} so that P(B/A) = P(A∩B)/(P(A))=(n(A∩B))/(n(A))=2/3. Illustration: In a college, 25% students failed in Mathematics, 155 students failed in Physics, and 10% failed in Mathematics and Physics. A student is selected at random: (i) If he failed in Physics, then find the chance of his failure in Mathematics, (ii) If he failed in Mathematics, then find the chance of his failure in Physics, (iii) Find the chance of his failure in Mathematics or Physics. Solution: Let E_{1} and E_{2} be the events of failure in mathematics and physics respectively. Let the total number of students appearing in the examination be 100. Since 25% students failed in Mathematics, n(E_{1}) = 25. => P(E_{1}) = n(E_{1})/(n(S))=25/100=1/4. Since 15% students failed in Physics, n (E_{2}) = 15 => P (E_{2}) = n (E_{2})/ (n(S)) =15/100=3/20. Again 10% students failed in Physics and Mathematics both, so n(E_{1}ÇE_{2}) = 10 => P (E_{1}∩E_{2}) = n (E_{1}∩E_{2} )/(n(S))=10/100=1/10.. (i) The chance of failure in Mathematics while he has failed in Physics is given by P (E_{1}/E_{2}) = = (P(E_{1}∩E_{2}))/P(E_{2} ) =(1/10)/(3/20)=2/3. (ii) The chance of failure in Physics while he has failed in Mathematics is given by P(E_{2}/E_{1}) = (P(E_{1}∩E_{2}))/P(E_{2} ) =(1/10)/(1/4)=2/5.. (iii) The chance of failure in Mathematics or Physics is: P (E_{1}UE_{2}) = P (E_{1}) + P (E_{2}) - P (E_{1}∩E_{2}) [Since both the events are not mutually exclusive] Conditional probability is an easy topic in the mathematics syllabus of the IIT JEE. Practice the questions on this topic to remain competitive in the JEE. Contact askiitians experts to get answers to your queries by filling up the form given below:

P (A/B) denotes the probability of event A happening, given that event B has happened. It is the conditional probability of A, given B. While calculating P (A/B), we assume that event B has occurred. It implies that the outcomes favorable to B become the total outcomes and hence outcomes favorable to P (A/B) are outcomes common to A and B.

P (A/B) = P (A given B)

= (Total no. of favorable cases)/ (Total no. of cases)

= n (A∩B)/n (B) = P (A∩B)/P (B).

Thus P (A/B) = P (A∩B)/P (B).

=> (P (A∩B) = P (B). (P (A/B) = P (A). (P(B/A)).

Note that if A and B are independent events then

P (A∩B) = P (A) P (B)

=> P (A/B) = P (A∩B)/P (B)

= P (A) P (B)/P (B)

= P (A)

which should be the case as occurrence of A does not depend upon B.

Similarly P (B/A) = P (B).

Sometimes it also happens so that the probability of one event is affected by occurrence of other events. Let a bag contains 3 red, 2 green and 4 yellow balls. A ball is drawn and found to be red. Consider the following cases:

(i) If the red ball is replaced then find the probability that the next ball to be drawn is yellow.

(ii) What happens when red ball will not be replaced?

Let, R: red ball is first drawn

Y: yellow ball is drawn next.

In first case when the red ball is replaced.

P(Y) = (Total number of yellow balls)/ (total number of balls) =4/9

In second case the red ball is not replaced, so,

The probability in these two cases is different. In the first case when the first ball was red and it replaced it does not affect the probability of yellow ball in second draw but in second case when the ball was not replaced the probability of yellow ball is changed. It happened because in second case the sample space is changed. So from here we can draw a conclusion that when occurrence of any event changes the sample space, the probability of other event change. Hence the probability also changes.

It means that, occurrence of any event change the sample space. Then Probability of other events changes, in other words probability of some event depends on the occurrence of other events. This is known as "Conditional Probability".

Let A and B be the two events. If A has already happened (i.e. given) then probability of B can be found by using the formula

P (B/A) = P (A∩B)/P (A)

If A and B are independent events, then

P (B/A) = P (B) i.e. P (A∩B) = P (A) P (B)

or, P(A/B) = P(A)

(i) n events A_{1}, A_{2}, A_{3}, ........... A_{n} are said to be pair-wise independent tiff P

(A_{i}∩A_{j}) = P(A_{i}) P(A_{j}), where 1, I = 1, 2, 3, ........., n

(ii) These n events are said to be mutually independent iff

P(A_{1} ∩ A_{2} ∩......... ∩ A_{n}) = P(A_{1}) P(A_{2}) ....... P(A_{n}).

Remark: If the set of n events related to a sample space are pair-wise independent, they must be mutually independent, but vice versa is not always true.

To apply total probability formula we must check whether A_{1} and A_{2} fulfill the following three conditions or not.

A_{1} ∩ A_{2} = Φ

and A_{1} U A_{2} = S.

A_{1 }∈ S and A_{2 }∈ S.

Suppose we have n events A_{1}, A_{2}, ........., A_{n} related to a sample space such that:

(i) They are mutually exclusive i.e.

A_{i} ∩ A_{j} = Φ, I, j = 1, 2 .......... n, I ≠ j.

(ii) They are exhaustive i.e.

A_{1} U A_{2} U A_{3} ......... U A_{n} = S.

(iii) They are proper subsets of sample space S i.e.

A_{i} S, i = 1, 2, .......... n.

Then probability of an event A which depends on the events A_{1}, A_{2}, ........., An can be calculated as

P (A) = ∑^{n}_{(i=1)} P(A/A_{i}) P(A_{i})

= P (A/A_{1}) P (A_{1}) + P (A/A_{2}) P (A_{2}) +......+ P (A/A_{n}) P (A_{n})

This is a formula for total probability.

Illustration:

Take the case of previous illustration in which a bag contains 3 Red, 2 Green and 4 Yellow balls. A ball is drawn and if it is Red or Yellow it is replaced otherwise not. Find the probability that second ball drawn is a yellow ball.

Solution:

First of all, we should be very clear about the event of which probability is to be found and the events on which it depends.

Let's A_{1}: First ball is either red or yellow

A_{2}: First ball is green

A: Second ball drawn is yellow

We have to calculate P(A) which definitely depends on A_{1} and A_{2}.

Therefore, P(A) = P(A/A_{1}) P(A_{1}) + P(A/A_{2}) P(A_{2})

P(A/A_{2}) = Probability of A when A_{1} is already happen.

= 4/9 (i.e. probability of A given A_{1})

P(A_{1}) = 7/9

P(A/A_{2}) = Probability of A when A_{2} is already happened

= (Probability of A given A_{2})

= 4/8

P(A_{2}) = 2/9

Hence, P(A) = 4/9×7/9+4/8×2/9 = 37/81

View this video for more on conditional probability

Example:

In a card game, suppose a player needs to draw two cards of the same suit in order to win. Of the 52 cards, there are 13 cards in each suit. Suppose first the player draws a heart. Now the player wishes to draw a second heart. Since one heart has already been chosen, there are now 12 hearts remaining in a deck of 51 cards. So the conditional probability

P (Draw second heart |First card was also a heart) = 12/51.

Suppose an individual applying to a college knows that there are 80% chances of his application being accepted and also there is a condition that only 60% of all the accepted students will be provided the dormitory housing. Hence, the chance of the student being accepted and receiving dormitory housing is given by

P (Accepted and Dormitory Housing)

= P (Dormitory Housing | Accepted) P (Accepted)

= (0.60)*(0.80) = 0.48.

A jar contains red and green toys. Two toys are drawn without replacement. The probability of selecting a red toy and then a green toy is 0.34, and the probability of selecting a red toy on the first draw is 0.47. What is the probability of selecting a green toy in the second draw, given that the first toy drawn was red?

Clearly it is a question of conditional probability. We need to find the probability of drawing a green toy having known that the first toy drawn was red.

P (green | red) = P (red and green) / P (red) = 0.34 / 0.47 = 0.72 = 72%.

At ABC Senior Secondary school, the probability that a student takes Science and English is 0.087. The probability that a student takes Science is 0.68. What is the probability that a student takes English given that the student is taking Science?

We need to find the probability of English given Science i.e.

P (English | Science) = P (Science and English) / P (Science) = 0.087 / 0.68 = 0.13 = 13%.

You may consult the Sample Papers to get an idea about the types of questions asked.

Remark: To calculate the probability of the intersection of more than two events, the conditional probabilities of all of the preceding events must be considered. In the case of three events, A, B, and C, the probability of the intersection P(A and B and C) = P(A)P(B|A)P(C|A and B).

Example: Consider the college applicant who has determined that he has 0.80 probability of acceptance and that only 60% of the accepted students will receive dormitory housing. Of the accepted students who receive dormitory housing, 80% will have at least one roommate. The probability of being accepted and receiving dormitory housing and having no roommates is calculated by: P(Accepted and Dormitory Housing and No Roommates) = P(Accepted)P(Dormitory Housing|Accepted)P(No Roomates|Dormitory Housing and Accepted) = (0.80)*(0.60)*(0.20) = 0.096. The student has about a 10% chance of receiving a single room at the college.

If a dice is thrown, what is the probability of occurrence of a number greater than 1, if it is known that only odd numbers can come up.

S = the sample space = {1, 2, 3, 4, 5, 6}

A = the event of occurrence of an odd number = {1, 3, 5}

B = the event of occurrence of a number greater than 1= {2, 3, 4, 5, 6}

Here A∩B = {3, 5}

so that P(B/A) = P(A∩B)/(P(A))=(n(A∩B))/(n(A))=2/3.

In a college, 25% students failed in Mathematics, 155 students failed in Physics, and 10% failed in Mathematics and Physics. A student is selected at random:

(i) If he failed in Physics, then find the chance of his failure in Mathematics,

(ii) If he failed in Mathematics, then find the chance of his failure in Physics,

(iii) Find the chance of his failure in Mathematics or Physics.

Let E_{1} and E_{2} be the events of failure in mathematics and physics respectively. Let the total number of students appearing in the examination be 100.

Since 25% students failed in Mathematics, n(E_{1}) = 25.

=> P(E_{1}) = n(E_{1})/(n(S))=25/100=1/4.

Since 15% students failed in Physics, n (E_{2}) = 15

=> P (E_{2}) = n (E_{2})/ (n(S)) =15/100=3/20.

Again 10% students failed in Physics and Mathematics both, so n(E_{1}ÇE_{2}) = 10

=> P (E_{1}∩E_{2}) = n (E_{1}∩E_{2} )/(n(S))=10/100=1/10..

(i) The chance of failure in Mathematics while he has failed in Physics is given by

P (E_{1}/E_{2}) = = (P(E_{1}∩E_{2}))/P(E_{2} ) =(1/10)/(3/20)=2/3.

(ii) The chance of failure in Physics while he has failed in Mathematics is given

by P(E_{2}/E_{1}) = (P(E_{1}∩E_{2}))/P(E_{2} ) =(1/10)/(1/4)=2/5..

(iii) The chance of failure in Mathematics or Physics is:

P (E_{1}UE_{2}) = P (E_{1}) + P (E_{2}) - P (E_{1}∩E_{2})

[Since both the events are not mutually exclusive] Conditional probability is an easy topic in the mathematics syllabus of the IIT JEE. Practice the questions on this topic to remain competitive in the JEE.

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