an um contains 9 red balls,7 white balls and 4 black balls.If two balls are drawn at random.find the probability that:

(i) both the balls are red

(ii)one ball is white

(iii)the ball are of the same colour

(iv)one is white and other red

(1) No. of ways of selecting both red balls = ^{9}C_{2}

Total no. of ways of selecting 2 balls from the urn =^{ (9 +7 + 4)}C_{2} = ^{20}C_{2}

Therefore, required probability = ^{9}C_{2} / ^{20}C_{2 }= 36 / 190 = **18 / 95.**

(2) No. of ways of selecting two balls in which one is white = Selecting one from white balls * Selecting one from remaining = ^{7}C_{1} * ^{(9 + 4)}C_{1}

Total no. of ways of selecting 2 balls from the urn =^{ (9 +7 + 4)}C_{2} = ^{20}C_{2}

Therefore, required probability = ^{7}C_{1} * ^{13}C_{1} / ^{20}C_{2} = **91 / 190.**

(3) No. of ways of selecting two balls of same color = Selecting both from red + Selecting both from white + Selecting both from black = ^{9}C_{2} + ^{7}C_{2} + ^{4}C_{2} = 63

Total no. of ways of selecting 2 balls from the urn =^{ (9 +7 + 4)}C_{2} = ^{20}C_{2}

Therefore, required probability = 63 / ^{20}C_{2} = **63 / 190.**

(4) No. of ways of selecting two balls such that one is white and one red = Selecting one from white balls * Selecting one from red balls = ^{7}C_{1} * ^{9}C_{1} = 63

Total no. of ways of selecting 2 balls from the urn =^{ (9 +7 + 4)}C_{2} = ^{20}C_{2}

Therefore, required probability = 63 / ^{20}C_{2} = **63 / 190.**

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