# 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 = 9C2

Total no. of ways of selecting 2 balls from the urn = (9 +7 + 4)C2 = 20C2

Therefore, required probability = 9C2 / 20C2 = 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 = 7C1 * (9 + 4)C1

Total no. of ways of selecting 2 balls from the urn = (9 +7 + 4)C2 = 20C2

Therefore, required probability = 7C1 * 13C1 / 20C2 = 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   =  9C2 + 7C2 + 4C2 = 63

Total no. of ways of selecting 2 balls from the urn = (9 +7 + 4)C2 = 20C2

Therefore, required probability = 63 / 20C2 = 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 = 7C1 * 9C1 = 63

Total no. of ways of selecting 2 balls from the urn = (9 +7 + 4)C2 = 20C2

Therefore, required probability = 63 / 20C2 = 63 / 190.

HOPE IE HELPED...

• 8
What are you looking for?