Compound Events,Conditional Probability

Compound Event

When two or more events occur in conjunction with each other in one or more random experiments the joint event is called compound event.

For example, if we draw a black ball and a white ball from a bag containing 4 black balls and 5 white balls,then the drawing of a black ball and a white ball is a compound event.

The compound events are two types they are independent and dependent events.

Independent Event

Two or more events are said to be independent events if the occurrence or non-occurrence of any one of them does not affect the occurrence of others.

For example, if a coin tossed two times,then getting the head in the second toss is independent of getting the head in the first toss.

Dependent Events

Two or more events are said to be dependent if the happening of any one of them affects (depend on ) the happening of others.If an event B occurs only after the occurrence of another event A, the event B is the dependent event and it is called conditional event of B given A which is denoted by \(\frac{B}{A}\) .

For example, a card is drawn from a well-shuffled pack of 52 cards . If a second card is drawn from the pack without replacing the first card in the pack, then the drawing of the second card depends on the drawing of the n first card.So, these drawing of the card is dependent events.

Conditional Probability

The probability of dependent events is called conditional probability.

Definition:

If A and B are two events in a sample space S such that B occurs only after the occurrence of A,then the probability of occurrence of B under the condition that A has already occurred ,denoted by P(\(\frac {B}{A}\)), is called conditional probability of B given A and it is given by,

$$P(\frac {B}{A})=\frac {n(A \cap B)}{n(A)}=\frac {P(A \cap B)}{P(A)}$$

provided that P(A) \(\neq\) 0

To find P(\(\frac {B}{A}\)), we need to consider only those sample points of B which lies in A and the sample points A. The sample points of B which lies in A are nthe sample points common to both A and B i.e. n(\(A \cap B\)).Since A has already occurred ,it folllows that A becomes a new sample space or reduced sample space for the conditional event \(\frac{B}{A}\).Therefore,

$$P(\frac {B}{A})=\frac {n(A \cap B)}{n(A)}=\frac{\frac{n(A \cap B)}{n(S)}}{\frac{n(A)}{n(S)}}=\frac {P(A \cap B)}{P(A)}$$

Hence, the conditional probability of B given A, P(\(\frac {B}{A}\)) is equal to the ratio of probability of occurrence of the compound event \(A \cap B\) to the probability of occurrence of the event A.

Similarly,the conditional probability of A given B is given by

$$P(\frac {A}{B})=\frac {P(A \cap B)}{P(B)} ;P(B)\neq 0$$

Theorem 10. For any three events A,B and C,

$$ i.\: \:P[\frac {(A \cup B)}{C}]=P(\frac {A}{C})+P(\frac {B}{C})-P[\frac {(A \cap B)}{C}] .......................(1)$$

$$ ii.\: \: P(A \cap \frac{B}{C})+P(A \cap \frac{ \bar B}{C})=P(\frac {A}{C}).....................(2)$$

Proof:

(i) This result follows from the distributive law and addition theorem of probability.

By addition theorem of probability we have for any two events A and B,

$$ P(A \cup B)=P(A) +P(B) -P(A \cap B)$$

For any three event A,B and C we have

$$(A \cap C) \cup (B \cap C)=(A \cup B) \cap C$$

$$ or \: \: P[ (A \cup B) \cap C]=P(A \cap C) +P(B \cap C)-P(A \cap B \cap C)$$

$$ or \; \; \frac {P[(A \cup B) \cap C]}{P(C)}= \frac {P(A \cap C)}{P(C)}+\frac {P(B \cap C)}{P(C)}-\frac {P(A \cap B \cap C)}{P(C)}$$

$$ or \; \; P[\frac {(A \cup B)}{C}] =P(\frac {A}{C})+P(\frac {B}{C})-P[\frac {(A \cap B)}{C}]$$

$$(ii) \; \; P(A \cap \frac {B}{C})+P(A \cap \frac{\bar B}{C})=\frac {P(A \cap B \cap C)}{P(C)}+ \frac{P(A \cap \bar B \cap C)}{P(C)}$$

$$=\frac {P(A \cap B \cap C) \cup P(A \cap \bar B \cap C)}{P(C)}$$

$$=\frac{P(A \cap C)}{P(C)}$$

$$=P(\frac {A}{C})$$

Example:

A card is drawn from a deck of 52 playing cards.What is the probability of drawing a king given the information that the drawn card is red?

Solution:

Here,the sample space S consist of 52 sample points, i.e. n(S)=52.

Let A be the event that the drawn card is red and B the event of drawing a king card.Then,

A={all 26 redd cards}

B={All 4 king cards}

and (\(A \cap B\))={2 red king cards}

$$Thus,P(A)= \frac {26}{52} ,P(B)=\frac {4}{52} \; \; and \; \;P(A \cap B)=\frac {2}{52}$$

Therefore the probability of drawing a king given that the drawn card is red is given by

$$ P(\frac {B}{A})=\frac {P(A \cap B)}{P(A)}=\frac{\frac{2}{52}}{\frac{26}{52}}=\frac{2}{26}=\frac{1}{13}$$

Independent Events

Definition: An event A is said to be statistically independent of another event B , if the conditional probability of A given B is equal to the unconditional probability of A, i.e. if

$$P(\frac {A}{B})=P(A)$$

Similarly, an event B is said to be statistically independent of event A if

$$P(\frac {B}{A})=P(B)$$

It is evident that if event A is independent of event B,then B is independent of A and also, we can say that A nad B are independent events.

Proof: $$Since \; \; \; A \cap \phi= \phi \; \; \; then$$

$$ P(A \cap \phi)=P(\phi)=0$$

$$Rightarrow \; \; P(A \cap \phi)=P(A).0=P(A) \times P(\phi)$$

Hence, A and \(\phi\) are independent events.

Again,since \(A \cap S\) =A then,

$$P(A \cap S)=P(A)=P(A).P(S) \; \; \; \; [P(S)=1]$$

hence ,A and S are independent.

Example:

Suppose two fair dice are thrown .Define two events A and B as follows:

A={First die shows an even number}

B={Second die shows a 5 or a 6}

Are A and B independent?

Solution

In throwing two fair dice ,the sample S consists of 36 wqually likely outcomes i.e. n(S)=36

Thus,

$$P(A)= \frac {n(A)}{n(S)}=\frac {18}{36}=\frac {1}{2}, \; \; \; \; P(B)=\frac {12}{36}=\frac {1}{3} \; \; and \; \; P(A \cap B)=\frac {6}{36}=\frac{1}{6}$$

$$ \therefore \; \; \; P( \frac {A}{B})=\frac {P(A \cap B)}{P(B)}=\frac{\frac{1}{6}}{\frac{1}{3}}=\frac {1}{2} \; \; \; and \; \; \; P(\frac {B}{A})=\frac {P(A \cap B)}{P(A)}=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac {1}{3}$$

$$ since \; \; P(\frac {A}{B})=P(A)=\frac {1}{2} \; \; \; and \; \; P(\frac {B}{A})=P(B)=\frac {1}{3}, \; \; the \; \; two \; \; events \; \; A \; \; and \; \; B \; \; are \; \; independent.$$