If the occurrence of one event does not affect the probability of the occurrence of the second event, then the events are independent.
For example, in rolling a die on two occasions, the outcome of the first throw would not affect the affect the probability of throwing a six as the second throw
Events are dependent when one even does affect the probability of the occurrence of the second
Definition: If two events A and B of a sample space S, these events are independent provided
P(A and B) = P (A ? B) = P(A) P(B)
If P(A ? B) ? P(A) P(B), then A and B are called dependent events
Proof: Suppose that the events A and B can happen in m1 and m2 cases respectively, and cannot
happen in n1 and n2 cases. Then by definition of probability
P(A) = m1/m1 + n1, P(B) = m2 / m2 + n2
Since the events A and B are independent, the number of cases in which both the events can happen simultaneously in m1m2 for from any one of m1 cases may be associated with any one of m2 cases
But the total number of ways which consists of the favourable and unfavourable cases for the simultaneous happening of A and B is (m1 + n1) (m2 + n2)
Hence, the probability P(A ? B) of simultaneous happening of A and B is given by
P(A ? B) = m1m2 /(m1 + n1)(m2 + n2) = m1/ (m1 + n1) ? m2 / (m2 + n2)
Hence the theorem
Illustrative problems on Compound Independent and Dependent Events
1) There are 4 girls and 2 boys one group and 3 girls and 7 boys in a second group. One child is selected at random from each group. Find the probability that a boy comes from the first group and a girl from the second
Solution: Let A be the event of selecting a boy from the first group, then P(A) = 2/6 = 1/3,
Let B be the event of selecting a girl from the second group, then P(B) = 3/10
... Required probability = P(A ? B) = P(A) P(B)
= 1/3 × 3/10 = 1/10
2) Three groups of children contain respectively 3 girls and 1 boy ; 2 girls and 2 boy; 1 girls and 3 boys. One child is selected at random from each group. Find the probability of selecting 1 girl and 2 boys.
Solution: One girl and two boys may be selected in the following independent ways:
First group Second group Third group
1 Girl Boy Boy
2 Boy Girl Boy
3 Boy Boy Girl
The probability of selection 1 is 3/4 × 2/4 × 3/4 = 9/32
The probability of selection 2 is 1/4 × 2/4 × 3/4 = 3/32
The probability of selection 3 is 1/4 × 2/4 × 1/4 = 1/32
Required probability = 9/32 + 3/32 + 1/32 = 13/32
Illustrative Problems on Compound Independent and Dependent Events
Solved Examples on Compound Independent and Dependent Events
Example1: A problem is given to three students A, B, C whose probabilities of solving it are 1/2, 1/3 and 1/4 respectively. Find probability that the problem will be solved.
Solution: The probability that A can solve the problem is 1/2 and the probability of not solving the problem is 1 – 1/2 = 1/2
Similarly, the probabilities of B and C of not solving the problem are 1- 1/3 = 2/3 and 1 – 1/4 = 3/4 respectively. Hence the probability of not solving the problem by all of them
= 1/2 × 2/3 × 3/4 = 1/4
Required probability of solving the problem = 1 – 1/4 = 3/4
Example2: There are 5 red and 4 white balls in a bag ad 3 green and 5 black balls in an other bag. One ball is drawn from each bag. Find the probability of drawing one white and one black ball
Solution: Total number of balls in the first bag = 5 + 4 = 9
Total number of balls in the second bag = 3 + 5 = 8
Let A be the event of drawing one white ball from the first bag, then P(A) = 4/9
Let B be the event of drawing black bag. from the second bag, then P(B) = 5/8
Required probability = P (drawing one white ball and one black ball)
= P(A and B) = P(A ? B) = P(A) P(B)
= (4/9) (5/8) = 5/18
Example3: Two bags contain respectively 3 white, 5 black and 5 white, 3 black balls. One ball is drawn from each bag. Find the probability that they are white.
Solution: Total number of balls in the first bag = 3 + 5 = 8
Total number of balls in the second bag = - 5 + 3 = 8
Let A be the event of drawing white ball from the first bag, then P(A) = 3/8
Let B be the event of drawing a white ball from the second bag, then P(B) = 5/8
... Required probability = P(A ? B)
= P(A). P(B) = 3/8 ? 5/8 = 15/64
Example4: There are two pack of card. One card is drawn from each pack at random. What is the probability of both being red?
Solution: Let A be the event of drawing one red card from the first pack. There are 26 red cards in each pack of 52 cards, then
P(A) = 26/52 = 1/2
Let B be the event of drawing one red card from the second pack, then P(B) = 26/52 = 1/2
Required probability = 1/2 × 1/2 = 1/4
Related Tags
Explain Compound Independent And Dependent Events , Introduction to Compound Independent And Dependent Events , What are Compound Independent And Dependent Events