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Two balls are picked in succesion without replacement -4 white balls and 5 green balls. Let Y be the random variable representing the number of green balls

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Let Y take the values 0,1, 2 where

p( Y=0) is the probability of picking zero green balls and picking two white balls

p( Y=1) is the probability of picking one green ball and one white ball.

p(Y=2) is the probability of picking two green balls and zero white balls.


we have a total of ( 5+4) balls = 9 balls, so there are 9C2 ways of picking 2 balls without replacement.

so, the above probabilities may be obtained as below


p( Y= 0) = ( ( 5C0 ) * (4C2) ) / ( 9C2 ) = 6/36 = 1/6


P( Y=1 ) = ( ( 5C1 ) * ( 4C1 ) ) / ( 9C2 ) = 20/36 =5/9


P( Y= 2) = ( ( 5C2 ) * ( 4C0) ) / ( 9C2 ) = 10/36 = 5/18


Hence we may construct the probability distribution representing the green balls that may be picked


y 0 1 2

p(y) 1/6 5/9 5/18


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