What is the probability distribution for Y,where Y is the number of red balls chosen. Bag A contains 2 red balls and 3 green balls.Two balls are chosen at random from the bag without replacement.Let X denote the number of red balls chosen.The probability distribution for X. When X=0,1,2 then P(X=x) is 3/10, 6/10, and 1/10. Bag B contains 4 red balls and 2 green balls.Two balls are chosen at random from bag B.

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

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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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