An urn contains 6 balls of which two are red and four are black. Two balls are drawn at random. Probability that they are of the different colours is
A. `(2)/(5)`
B. `(1)/(15)`
C. `(8)/(15)`
D. `(4)/(15)`
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)) =...
2 Answers 1 viewsNumber of ways of drawing 7 balls (second draw) = `.^(10)C_(7)` For each set of 7 balls of the second draw, 3 must be common to the set of 5...
2 Answers 3 viewsThe 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,...
2 Answers 1 viewsLet A nad B denote, respectively, the events that first ball drawn is black and second ball drawn in white. We have to find `P(AnnB) or P(AB).` ` P(AnnB)P(A)P(B//A)` Now,...
2 Answers 1 viewsProcedure 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...
2 Answers 1 viewsLet `A_(i)` denote the event that the number I appears on the dice and let E denote the event that only white balls are drawn. Then `P(A_(i))=1/6"for"i=1,2..,6` `and P(E//A_(i))=(""^(6)C_(i))/(""^(10)C_(i)),i=,2,.., 6`...
2 Answers 1 viewsCorrect Answer - `1/7` Required probability `=P(WBWB)+P(BWBW)` `=P(W)P(B//W)P(W//WB)P(B//BWB)` `=5/8xx3/7xx4/6xx2/5+3/8xx5/7xx2/6xx4/5` `=(1)/(14)+(1)/(14)=1/7`
2 Answers 1 viewsCorrect Answer - D Let A nad B, respectively, be the events that urn A and urn B are selected.Let R be the event that the selected ball is red. Since...
2 Answers 1 viewsCorrect Answer - D In the first case, the urn contains 3 red and n white balls. The probability that color of both the balls matches is `(""^(3)C_(2)+""^(n)C_(2))/(""^(n+3)C_(2))=1/2` `or(6+n(n-1))/((n+3)(n+2))=1/2` `or2(n^(2)-n+6)=n^(2)+5n+6` `orn^(2)-7n+6=0`...
2 Answers 1 viewsCorrect 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`
2 Answers 1 views