It is given that Bag B contains 4 red balls and 2 green balls. Two balls are chosen at random from bag B. The probability distribution of Y, where Y is the number of red balls chosen is:

0: (2/6)*(1/5) = 2/30

1: 2*(4/6)*(2/5) = 16/30

2: (4/6)*(3/5) = 12/30

**The required probability distribution of the number of red balls chosen is 0: 2/30, 1: 16/30 and 2: 12/30**

The problem provides the information that the bag B contains 4 red balls and 2 green balls.

Selecting two balls at random from bag B, the probability to select two greens is of `4/6*3/5 = 12/30` .

Selecting two balls at random from bag B, the probability to select two reds is of `2/6*1/5 = 2/30` .

**Hence, evaluating the probability distribution for the number of red balls selected yields: if y=0 then `P(Y = 0) = 2/30` and if y=2, then `P(Y=2) = 12/30` .**

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