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An um contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red. then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is

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26/49

32/49

21/49

27/49

answer is B.

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Detailed Solution

E₁ : Event of drawing a Red ball and placing a green ball in the bag $\Rightarrow$ P (E₁) = 5/7

E₂ : Event of drawing a green ball and placing a red ball in the bag ) $\Rightarrow$ P (E₂) =2/7

E: Event of drawing are red ball in second draw P(E|E₁) = 4/7, P(E|E₂) = 6/7

$P (E) = P (E_{1}) \times P (E ∣ E_{1}) + P (E_{2}) \times P (E ∣ E_{2}) = \frac{5}{7} \times \frac{4}{7} + \frac{2}{7} \times \frac{6}{7} = \frac{32}{49}$

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