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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 âˆ© B = {HTT}

P (A âˆ© 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 âˆ© B"

P (A|B) = Number of elementary events favourable to A âˆ© B / Number of elementary events favourable to B

= n(A âˆ© B)/n(B)

= (n(A âˆ© B)/n(S))/(n(B)/n(S))

Probability of A, given that B has already occurred = P (A|B) = P(A âˆ© 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 âˆ© B)/P(B), P (B) â‰ 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 âˆ© B = {HTT}

P (A âˆ© 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 âˆ© A)/P(A)

= P(A)/P(B)

= 1

P (A|A) = P(A âˆ© 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) â‰ 0, then P((A âˆª B)|E) = P(A|E) + P(B|E) - P((A âˆ© B)|E).

LHS = P ((A âˆª B) |E)

= P[(A âˆª B) âˆ© E]/P(E)

= P[(A âˆ© E) âˆª (B âˆ© E)]/P(E)

(by distributive law of union of sets over intersection)

= (P(A âˆ© E) + P(B âˆ© E) - P(A âˆ© B âˆ© E))/P(E) [âˆµ P(X âˆª Y) = P(X) + P(Y) - P(X âˆ© Y)]

= P(A âˆ© E)/P(E) + P(B âˆ© E)/P(E) - (P(A âˆ© B) âˆ© E)/P(E)

= P(A|E) + P(B|E) - P((A âˆ© B)|E) = RHS.

Hence, P((A âˆª B)|E) = P(A|E) + P(B|E) - P((A âˆ© B)|E)

For two disjoint events A and B, A âˆ© B = âˆ…

â‡’ P((A âˆ© B)|E) = 0

P((A âˆª 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)

â‡’ P(A âˆª A'|B) = 1 (âˆµ A âˆª A' = S)

â‡’ P(A|B) + P (Aâ€²|B) = 1 (âˆµ A and A' are disjoint events)

â‡’ P (Aâ€²|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 âˆ© B = {HTT}

P (A âˆ© 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 âˆ© B"

P (A|B) = Number of elementary events favourable to A âˆ© B / Number of elementary events favourable to B

= n(A âˆ© B)/n(B)

= (n(A âˆ© B)/n(S))/(n(B)/n(S))

Probability of A, given that B has already occurred = P (A|B) = P(A âˆ© 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 âˆ© B)/P(B), P (B) â‰ 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 âˆ© B = {HTT}

P (A âˆ© 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 âˆ© A)/P(A)

= P(A)/P(B)

= 1

P (A|A) = P(A âˆ© 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) â‰ 0, then P((A âˆª B)|E) = P(A|E) + P(B|E) - P((A âˆ© B)|E).

LHS = P ((A âˆª B) |E)

= P[(A âˆª B) âˆ© E]/P(E)

= P[(A âˆ© E) âˆª (B âˆ© E)]/P(E)

(by distributive law of union of sets over intersection)

= (P(A âˆ© E) + P(B âˆ© E) - P(A âˆ© B âˆ© E))/P(E) [âˆµ P(X âˆª Y) = P(X) + P(Y) - P(X âˆ© Y)]

= P(A âˆ© E)/P(E) + P(B âˆ© E)/P(E) - (P(A âˆ© B) âˆ© E)/P(E)

= P(A|E) + P(B|E) - P((A âˆ© B)|E) = RHS.

Hence, P((A âˆª B)|E) = P(A|E) + P(B|E) - P((A âˆ© B)|E)

For two disjoint events A and B, A âˆ© B = âˆ…

â‡’ P((A âˆ© B)|E) = 0

P((A âˆª 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)

â‡’ P(A âˆª A'|B) = 1 (âˆµ A âˆª A' = S)

â‡’ P(A|B) + P (Aâ€²|B) = 1 (âˆµ A and A' are disjoint events)

â‡’ P (Aâ€²|B) = 1 - P (A|B).