Notes On Conditional Probability - CBSE Class 12 Maths
The set of all the possible outcomes of a random experiment is called sample space. It is denoted by S. Each outcome in a sample space is called a sample point. In the sample space for the experiment when three fair coins are tossed. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Here, each sample point is equally likely. Probability of each sample point = 1/8 Let A be an event that at least two tails appear. That is, two or more than two tails appear. A = {HTT, THT, TTH, TTT} P (A) = P ({HTT}) + P ({THT}) + P ({TTH}) + P ({TTT}) = 1/8 + 1/8 + 1/8 + 1/8 = 1/2 Let B is event first coin shows head. B = {HTT, HHT, HTH, HHH} P (B) = P ({HTT}) + P ({HHT}) + P ({HTH}) + P ({HHH}) = 1/8 + 1/8 + 1/8 + 1/8 = 1/2 S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} A = {HTT, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} A ∩ B = {HTT} P (A ∩ B) = P ({HTT}) = 1/8 Let's find the probability of A (at least two tails), if event B has already occurred. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} Now, sample space for event A is the set of all favourable occurrences of event B. Sample space for event A = {HTT, HHT, HTH, HHH} Event A: At least two tails The sample point of B favourable to the occurrence of event A is HTT. A = {HTT} The sample points in B are equally likely. Probability of each sample point in B = 1/4 P (A) = P ({HTT}) = 1/4 Conditional probability The probability of event A is called the conditional probability of A given that event B has already occurred. This is denoted by P (A|B) and read as probability of A given B. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} A = {HTT, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} {HTT} is common to events A and B. The outcomes of event A = The sample points of event "A ∩ B" P (A|B) = Number of elementary events favourable to A ∩ B / Number of elementary events favourable to B = n(A ∩ B)/n(B) = (n(A ∩ B)/n(S))/(n(B)/n(S)) Probability of A, given that B has already occurred = P (A|B) = P(A ∩ B)/P(B) If A and B are two events of same sample space S, then the conditional probability of event A given that event B has occurred, is given by P (A|B) = P(A ∩ B)/P(B), P (B) ≠ 0 S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} A = {HTT, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} Let's find the probability of A, if event B has already occurred. A ∩ B = {HTT} P (A ∩ B) = 1/8 P (B) = 1/2 Probability of A, given that B has already occurred = P (A|B) = (1/8)/(1/2) = 1/4 The definition of conditional probability can also be applied in a general case, where the elementary events are not equally likely. Properties of Conditional Property Property: If A is an event associated with sample space S, then P (S|A) = P (A|A) = 1. P (S|A) = P(S ∩ A)/P(A) = P(A)/P(B) = 1 P (A|A) = P(A ∩ A)/P(A) = P(A)/P(A) = 1 Hence, P (S|A) = P (A|A) = 1 Property: If A and B are two events of sample space S, and E is an event of S such that P(E) ≠ 0, then P((A ∪ B)|E) = P(A|E) + P(B|E) - P((A ∩ B)|E). LHS = P ((A ∪ B) |E) = P[(A ∪ B) ∩ E]/P(E) = P[(A ∩ E) ∪ (B ∩ E)]/P(E) (by distributive law of union of sets over intersection) = (P(A ∩ E) + P(B ∩ E) - P(A ∩ B ∩ E))/P(E)  [∵ P(X ∪ Y) = P(X) + P(Y) - P(X ∩ Y)] = P(A ∩ E)/P(E) + P(B ∩ E)/P(E) - (P(A ∩ B) ∩ E)/P(E) = P(A|E) + P(B|E) - P((A ∩ B)|E) = RHS. Hence, P((A ∪ B)|E) = P(A|E) + P(B|E) - P((A ∩ B)|E) For two disjoint events A and B, A ∩ B = ∅ ⇒ P((A ∩ B)|E) = 0 P((A ∪ B)|E) = P(A|E) + P(B|E), if A and B are disjoint sets. Property: If A and B are two events of sample space S, then P(A'|B) = 1 - P(A|B). We know that P(S|B) = 1 (From property I) ⇒ P(A ∪ A'|B) = 1 (∵ A ∪ A' = S) ⇒ P(A|B) + P (A′|B) = 1 (∵ A and A' are disjoint events) ⇒ P (A′|B) = 1 - P (A|B).

#### Summary

The set of all the possible outcomes of a random experiment is called sample space. It is denoted by S. Each outcome in a sample space is called a sample point. In the sample space for the experiment when three fair coins are tossed. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Here, each sample point is equally likely. Probability of each sample point = 1/8 Let A be an event that at least two tails appear. That is, two or more than two tails appear. A = {HTT, THT, TTH, TTT} P (A) = P ({HTT}) + P ({THT}) + P ({TTH}) + P ({TTT}) = 1/8 + 1/8 + 1/8 + 1/8 = 1/2 Let B is event first coin shows head. B = {HTT, HHT, HTH, HHH} P (B) = P ({HTT}) + P ({HHT}) + P ({HTH}) + P ({HHH}) = 1/8 + 1/8 + 1/8 + 1/8 = 1/2 S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} A = {HTT, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} A ∩ B = {HTT} P (A ∩ B) = P ({HTT}) = 1/8 Let's find the probability of A (at least two tails), if event B has already occurred. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} Now, sample space for event A is the set of all favourable occurrences of event B. Sample space for event A = {HTT, HHT, HTH, HHH} Event A: At least two tails The sample point of B favourable to the occurrence of event A is HTT. A = {HTT} The sample points in B are equally likely. Probability of each sample point in B = 1/4 P (A) = P ({HTT}) = 1/4 Conditional probability The probability of event A is called the conditional probability of A given that event B has already occurred. This is denoted by P (A|B) and read as probability of A given B. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} A = {HTT, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} {HTT} is common to events A and B. The outcomes of event A = The sample points of event "A ∩ B" P (A|B) = Number of elementary events favourable to A ∩ B / Number of elementary events favourable to B = n(A ∩ B)/n(B) = (n(A ∩ B)/n(S))/(n(B)/n(S)) Probability of A, given that B has already occurred = P (A|B) = P(A ∩ B)/P(B) If A and B are two events of same sample space S, then the conditional probability of event A given that event B has occurred, is given by P (A|B) = P(A ∩ B)/P(B), P (B) ≠ 0 S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} A = {HTT, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} Let's find the probability of A, if event B has already occurred. A ∩ B = {HTT} P (A ∩ B) = 1/8 P (B) = 1/2 Probability of A, given that B has already occurred = P (A|B) = (1/8)/(1/2) = 1/4 The definition of conditional probability can also be applied in a general case, where the elementary events are not equally likely. Properties of Conditional Property Property: If A is an event associated with sample space S, then P (S|A) = P (A|A) = 1. P (S|A) = P(S ∩ A)/P(A) = P(A)/P(B) = 1 P (A|A) = P(A ∩ A)/P(A) = P(A)/P(A) = 1 Hence, P (S|A) = P (A|A) = 1 Property: If A and B are two events of sample space S, and E is an event of S such that P(E) ≠ 0, then P((A ∪ B)|E) = P(A|E) + P(B|E) - P((A ∩ B)|E). LHS = P ((A ∪ B) |E) = P[(A ∪ B) ∩ E]/P(E) = P[(A ∩ E) ∪ (B ∩ E)]/P(E) (by distributive law of union of sets over intersection) = (P(A ∩ E) + P(B ∩ E) - P(A ∩ B ∩ E))/P(E)  [∵ P(X ∪ Y) = P(X) + P(Y) - P(X ∩ Y)] = P(A ∩ E)/P(E) + P(B ∩ E)/P(E) - (P(A ∩ B) ∩ E)/P(E) = P(A|E) + P(B|E) - P((A ∩ B)|E) = RHS. Hence, P((A ∪ B)|E) = P(A|E) + P(B|E) - P((A ∩ B)|E) For two disjoint events A and B, A ∩ B = ∅ ⇒ P((A ∩ B)|E) = 0 P((A ∪ B)|E) = P(A|E) + P(B|E), if A and B are disjoint sets. Property: If A and B are two events of sample space S, then P(A'|B) = 1 - P(A|B). We know that P(S|B) = 1 (From property I) ⇒ P(A ∪ A'|B) = 1 (∵ A ∪ A' = S) ⇒ P(A|B) + P (A′|B) = 1 (∵ A and A' are disjoint events) ⇒ P (A′|B) = 1 - P (A|B).

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