Dependent Events in Probability

If the occurrence of event A affects the occurrence or non-occurrence of event B, the events are termed as dependent events. For example: A coloured ball is drawn from a bag. If another ball is drawn from the bag before replacing the first ball, the probability of drawing the second ball will be affected by the probability of drawing the first ball. If the first ball was replaced, the events would have been independent. To find the probability of two dependent events occurring simultaneously, conditional probability is used. If one is altered, it will definitely affect the probability of the other event.

Conditional Probability

The probability of an event given that another event has occurred is termed as conditional probability. If A and B are two events, then the conditional probability of A given that event B has occurred is given by,

P(A/B) = \[\frac{P(A∩B)}{P(B)}\]

Similarly, if A and B are two events, then the conditional probability of B given that event A has occurred is given by,

P(B/A) = \[\frac{P(A∩B)}{P(A)}\]

Theorem

Assuming A and B to be two dependent events then,

P(A∩B) = P(A).P(B/A)

The probability of simultaneous happening of two events A and B is equal to the probability of A multiplied by the conditional probability of B with respect to A.

Similarly,

P(A∩B) = P(B).P(A/B)

The probability of simultaneous happening of two events A and B is equal to the probability of B multiplied by the conditional probability of A with respect to B.

Solved Examples

Example 1:

There are 3 red, 6 white and 7 blue balls in a bag. If two balls are drawn one by one, find the probability that the first ball is white and the second ball is blue when the first ball drawn is not replaced.

Solution:

Let A be the event of drawing a white ball and B be the event of drawing second a blue ball. Since, the first ball is not replaced before drawing the second ball, the two events are dependent.

Total number of balls = 3 + 6 + 7 = 16

Number of white balls = 6

Therefore,

Probability of drawing a white ball, P(A) = \[\frac{6}{16}\]

The number of balls in the bag is now 16 - 1 = 15

Number of blue balls = 7

A blue ball is drawn given that a white ball is drawn. Therefore, conditional probability of B given that A has occurred is,

P(B/A) = \[\frac{7}{15}\]

Now, the probability that events A and B occur simultaneously is given by,

P(A∩B) = P(A).P(B/A)

Substituting the respective values, 

P(A∩B) = \[\frac{6}{16}\] × \[\frac{7}{15}\] = \[\frac{7}{40}\]

Therefore, the probability that the first ball is white and the second ball is blue when the first ball drawn is not replaced is 7/40.

Example 2:

Two cards are drawn one by one from a pack of 52 cards without replacement. What is the probability that the first card drawn is a king and second is queen?

Solution:

Let A be the event of drawing a king and B be the event of drawing a queen. Since, the first card, that is, king is not replaced before drawing the second card, that is queen, the two events are dependent.

Total number of balls = 52

Number of kings = 4

Therefore,

Probability of drawing a king, P(A) = \[\frac{4}{52}\]

The number of cards in the deck now is 52 - 1 = 51

Number of queen = 4

A queen is drawn given that a king is drawn. Therefore, conditional probability of B given that A has occurred is,

P(B/A) = \[\frac{4}{51}\]

Now, the probability that events A and B occur simultaneously is given by,

P(A∩B) = P(A).P(B/A)

Substituting the respective values, 

P(A∩B) = \[\frac{4}{52}\] × \[\frac{4}{51}\] = \[\frac{4}{663}\]

Therefore, the probability that the first card drawn is a king and second is queen is 4/663.

Example 3:

There are 19 tickets in a bag numbered from 1 to 19. One ticket is dawn and then a second ticket is drawn without replacement. What is the probability that both the tickets will show even number?

Solution:

Let A be the event of getting an even number in the first draw and B be the event of getting an even number in the second draw. Since, the second ticket is drawn without replacing the first ticket, the events are dependent.

Total number of tickets = 19

Even numbered tickets = 9

Therefore,

Probability of getting an even number in the first draw = \[\frac{9}{19}\]

Number of tickets left = 18

Number of even numbered tickets left = 8

An even numbered card is drawn given that an even numbered card is drawn before. Therefore, conditional probability of B given that A has occurred is,

P(B/A) = \[\frac{8}{18}\]

Now, the probability that events A and B occur simultaneously is given by,

P(A∩B) = P(A).P(B/A)

Substituting the respective values, 

P(A∩B) = \[\frac{9}{19}\] × \[\frac{8}{18}\] = \[\frac{4}{19}\]

Therefore, the probability that both the tickets will show even number is 4/663.

Did You Know

• If the two events are independent, that is the occurrence of one event does not affect the occurrence or non-occurrence of another event, then the probability of the two events occurring simultaneously is the product of their respective probabilities.

• For conditional probability of event A with respect to event B, probability of event B can never be zero. Conversely, for conditional probability of event B with respect to event A, probability of event A can never be zero.

• If two events can never occur simultaneously, they are termed as mutually exhaustive events, that is A∩B= ф.

• The formula for finding the probability of two events occurring simultaneously is derived from the multiplication theorem of probability.

• A∩B is represented by the intersection of two sets in a Venn diagram.

• The set of all possible outcomes in an experiment is termed as sample space. The sample space of an experiment is affected if the events are dependent.