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An urn contains 5 red balls and 2 green balls. A ball is drawn. If it’s green, a red ball is added to the urn, and if it’s red, a green ball is added to the urn. (The original ball is not returned to the urn). Then a second ball is drawn. What is the probability that the second ball is red? |
Answer» A: Event of drawing a red ball and placing a green ball in the urn B: Event of drawing a green ball and placing a red ball C: Event of drawing a red ball in the second draw P(A) = 5/7 P(B) = 2/7 P(C/A) = 4/7 P(C/B) = 6/7 Required probability P(C) = P(A) P(C/A) + P(B) P(C/B) \(\frac {5}{7} \times \frac {4}{7} +\frac{2}{7}\times\frac{6}{7}\) = 32/49 |
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