an urn contains 6 red 5 green 2 blue and 7 yellow ball one b

an urn contains 6 red, 5 green, 2 blue and 7 yellow ball. one ball is drawn and replaced. a second ball is the probability the first is green and the second is red?

three ball are drwn. what is the probability at least one is red?

Solution

Bag 1 contains 5 red and 3 green balls.
Bag 2 contains 4 red and 6 green balls.
Probability = No.of favorable outcomes/Total No.of outcomes
Probability of selecting a red ball from bag 1
P(R1) = 5C1/8C1 = 5/8
Probability of selecting a green ball from bag 1
P(G1) = 3C1/8C1 = 3/8
Probability of selecting a green ball from bag 2
P(G2) = 6C1/10C1 = 6/10 = 3/5
Probability of selecting a red ball from bag 2
P(R2) = 4C1/10C1 = 4/10 = 2/5
Probability of selecting a red ball from bag 1 and green ball from bag 2
As both the events are independent, the probability of occurrence of both the events is P(T1) = P(R1) x P(G2)
         = 5/8 x 3/5
         = 3/8
Therefore the probability is 3/8.
Probability of selecting a green ball from bag 1 and red ball from bag 2
P(T2) = P(R2) x P(G1)
           = 2/5 x 3/8
           = 6/40
Total probability = P(T1) + P(T2)
                              = 3/8 + 6/40
                               = 21/40