Let `n_(1)and n_(2)` be the number of red and black balls, respectively, in box I. Let `n_(3) and n_(4)` be the numbers of red and black balls, respec

Bag I : 3 red balls and 0 white ball Bag II : 2 red balls and 1 white ball. Bag III : 0 red ball andf 3 white balls....

Let `U_(1)`={2 white, 3 black balls} `U_(2)`={3 white , 2 black balls} and `U_(3)`={4 white ,1 black balls} `thereforeP(U_(1))=P(U_(2))=P(U_(3))=1/3` Let `E_(1)` be the event that a ball is chosen from...

Probability of drawing 2 green balls and one blue ball `=P_(G)cdotP_(G)cdotP_(B)+P_(B)cdotP_(G)cdotP_(G)+P_(G)cdotP_(B)cdotP_(G)` `=3/8cdot2/7cdot2/6+2/8cdot3/7cdot2/6+3/8cdot2/7cdot2/6` `=1/28+1/28+1/28=3/28`

To draw 2 black, 4 white, and 3 red balls in order is same as arranging two black balls at first 2 places, 4 white balls at next 4 places,...

Correct Answer - A::B Let the number of red and blue balls be r and b, respectively. Then, the probability of drawing two red balls is `p_(1) = (.^(r )C_(2))/(.^(r+b)C_(2)) =...

The urn contains 5 red and 2 black balls.  If two balls are drawn from the urn, it contains either 0 or 1 or 2 black balls.  X can take values 0,...

Procedure of drawing th balls has to end at the rth draw. So, exactly two black balls are drawn in first (r-1) draws and third black ball is drawn in...

Correct Answer - C In any trial P (getting white ball) = P (W) =1/2 P(getting black ball)`=P(B)=1//2` Required event will occur if in the first six trial, 3 white balls...

Correct Answer - A P (required) = P (all are white) + P (all are red) + P (all are black) `=1/6xx2/9xx3/12+3/6xx3/9xx4/12+2/6xx4/9xx2/12=6/648+36/648+40/648=82/648`

Correct Answer - B `(b)` Let `A` be the event of drawing a black ball. `A_(1)`, the event of choosing the first box. `A_(2)`, the event of choosing the second box....