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).
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).