A bag contains 4 red and 4 black balls, another bag contains 2 red and

1.	A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
Answer» Let E₁ : first bag is selected, E₂ : second bag is selected Then, E₁ and E₂ are mutually exclusive and exhaustive. Moreover, P(E₁) = P(E₂) = 1/2 Let E : ball drawn is red. P(E/E₁) = P(drawing a red ball from first bag) = 4/8 = 1/2 P(E/E₂) = P(drawing a red ball from second bag) = 2/8 = 1/4 By using Baye’s theorem, Required probability = P(E₁/E) = (P(E₁)P(E/E₁))/(P(E₁)P(E/E₁) + P(E₂)P(E/E₂)) = (1/2 x 1/2)/(1/2 x 1/2 + 1/2 x 1/4) = (1/4)/(1/4 + 1/8) = (1/4)/(2 + 1)/8 = (1/4)/(3/8) = 2/3

A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Answer»

Let E₁ : first bag is selected, E₂ : second bag is selected

Then, E₁ and E₂ are mutually exclusive and exhaustive. Moreover,

P(E₁) = P(E₂) = 1/2

Let E : ball drawn is red.

P(E/E₁) = P(drawing a red ball from first bag) = 4/8 = 1/2

P(E/E₂) = P(drawing a red ball from second bag) = 2/8 = 1/4

By using Baye’s theorem,

Required probability = P(E₁/E)

= (P(E₁)P(E/E₁))/(P(E₁)P(E/E₁) + P(E₂)P(E/E₂))

= (1/2 x 1/2)/(1/2 x 1/2 + 1/2 x 1/4) = (1/4)/(1/4 + 1/8)

= (1/4)/(2 + 1)/8 = (1/4)/(3/8) = 2/3

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