there are x green beads in a pocket .the number of red beads is x/2
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Answer:
A bag contains 5 red and 3 green balls. Another bag contains 4 red and 6 green balls. If one ball is drawn from each bag what is the probability that one ball is red and one is green?
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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
Bag 1 contains 8 balls
Probability of selecting a red ball from bag 1
P(R1) = 5C1(5 red balls)/8C1 = 5/8
Probability of selecting a green ball from bag 1
P(G1) = 3C1(3 green balls)/8C1 = 3/8
Bag 2 contains 10 balls.
Probability of selecting a green ball from bag 2
P(G2) = 6C1(6 green balls)/10C1 = 6/10 = 3/5
Probability of selecting a red ball from bag 2
P(R2) = 4C1(4 red balls)/10C1 = 4/10 = 2/5
We have 2 possibilities here,
1. Probability of drawing a red ball from bag 1 and green ball from bag 2
(PT1) = P(R1) x P(G2)
= 5/8 x 3/5
= 3/8
We are multiplying here because, the selection of red balls does't depend on green balls and vice versa,i.e both are independent.
2. Probability of drawing 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
To get the overall probability, we take the sum of individual probabilities.
Total probability = P(T1) + P(T2)
= 3/8 + 6/40
= 21/40