One box contains six red balls and four green balls, and a second box contains s

Question

One box contains six red balls and four green balls, and a second box contains seven red balls and three green balls. A ball is randomly chosen from the first box and placed in the second box. Then a ball is randomly selected from the second box and placed in the first box. a. What is the probability that a red ball is selected from the first box and a red ball is selected from the second box? b. At the conclusion of the selection process, what is the probability that the numbers of red and green balls in the first box are identical to the numbers at the beginning?

Explanation / Answer

a. Probability of selecting a Red ball from the first Box = 6/10 = 0.6

Probability of selecting a Red ball from the second Box =

Pr(Red is chosen from first Box) * Pr(Red is chosen from second Box) + Pr( Green s chosen from the first Box) * (Red is chosen from second Box) = 6/10 * 8/11 + 4/10 * 7/11 = 76/110 = 38/55

Here the term 8/11 will come as there will be 11 balls now in box 2 and out of which 7 +1 = 8 are red now. Similarly, in next part there are 11 balls, out of which 7 are red because we have chosen a green ball from Box 1.

b. Number of Red balls and green balls in first box are identical after the reshuffling.

It wil be possible if the ball chosen from Box 2 shall be of same color of ball chosen from Box 1.

so Pr(Red and Green balls are same as earlier) = Pr(Red from first) * Pr( Red from Second) + Pr( Green From first) * Pr( Green from second)

= 6/10 * 8/11 + 4/10* 4/11 = 32/55

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