Question

In a bag there are 6 red balls and 4 white ball and in another bag there are 5red balls and 6white balls
A)what is the probability of taking white ball from the first bag?
B)which bag has more probability of getting a red ball?

Answer 1

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Bag 1 contains 6 red and 7 white balls.

Bag 2 contains 5 red and 3 white balls.

Probability = No.of favorable outcomes/Total No.of outcomes

Bag 1 contains 13 balls

Probability of selecting a red ball from bag 1

P(R1) = 6C1(6 red balls)/13C1 = 6/13

Probability of selecting a white ball from bag 1

P(G1) = 7C1(7 white balls)/13C1 = 7/13

Bag 2 contains 8 balls.

Probability of selecting a white ball from bag 2

P(G2) = 3C1(3 white balls)/8C1 = 3/8

Probability of selecting a red ball from bag 2

P(R2) = 5C1(5 red balls)/8C1 = 5/8

We have 2 possibilities here,

1. Probability of drawing a red ball from bag 1 and white ball from bag 2

(PT1) = P(R1) x P(W2)

= 6/13 x 3/8

= 18/104

We are multiplying here because, the selection of red balls does't depend on white balls and vice versa,i.e both are independent.

2. Probability of drawing a white ball from bag 1 and red ball from bag 2

P(T2) = P(R2) x P(W1)

= 5/8 x 7/13

= 35/104

To get the overall probability, we take the sum of individual probabilities.

Total probability = P(T1) + P(T2)

= 18/104 + 35/104

= 53/104

Answer 2

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