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A Bag Contains 8 Red And 7 Black Balls Two Balls Are Drawn At Random The Probability That Both The Balls Are Of The Same Colour Is

1) 14/15 2) 11/15 3) 7/15 4) 4/15 Solution: Option (3) 7/15 Total cases = 15C2 = 105 Favourable cases = 8C2+7C2 = 28+21 = 49... View Article

1) 21/316 2) 19/316 3) 1/4 4) None of these Solution: Option (2) 19/316. Total cases = 80C2 Favourable cases = 20C2 (As, there... View Article

1) 1/15 2) 14/15 3) 1/5 4) 4/5 Solution: Option (4) 4/5 Given that, S = {1, 2,3, 4, 5, 6} Total cases = 2!. 6C2 = 2(15) = 30... View Article

1) 2/19 2) 3/29 3) 17/19 4) 4/19 Solution: Option (3) 17/19 Total cases = 20C3 From the first 20 integers, 10 integers should be... View Article

1) 14/29 2) 16/29 3) 15/29 4) 10/29 Solution: Option (3) 15/29 Total cases = 30C2 = 435 The sum is odd, if one of the integers is... View Article

1) 1/3 2) 1/4 3) 5/12 4) 2/3 Solution: Option (4) ⅔ Total cases = 12C1 = 12 Favourable cases = 5C1 + 3C1 = 5+3 = 8 Hence, the... View Article

1) (8C6 . 7C5)/15C11 2) (8C6 + 7C5)/15C11 3) 15/28 4) None of these Solution: Option (1) (8C6 . 7C5)/15C11 Total cases = 15C11... View Article

1) 2/801 2) 2/623 3) 1/267 4) 1/623 Solution: Option (1) 2/801 Total cases = 90C5 Favourable cases = 88C3 (The tickets drawn are... View Article

1) 1/34 2) 1/35 3) 1/17 4) 1/68 Solution: Option (2) 1/35 Total cases = 7! Ways Favourable cases = 4!. 3! (As boys can be arranged... View Article

1) 38C3 2) 37C2 3) 37C2/38C3 4) 666/8436 Solution: Option (3) 37C2/38C3 Total cases = 38C3 Favourable cases = 37C2 Hence,... View Article

1) 4/5 2) 3/5 3) 2/5 4) 1/5 Solution: Option (1) 4/5 Total number of ways = 10! ways. Two students can be seated side by side = 2.... View Article

1) 1/8 2) 2/21 3) 1/32 4) 1/35 Solution: Option (4) 1/35 In an octagon, there are 8 vertices. Out of which 4 vertices are chosen.... View Article

1) 3/11 2) 2/11 3) 8/11 4) None of these Solution: Option (1) 3/11 Total cases = 12C3 Favourable cases = 3C1. 4C1. 5C1 Hence,... View Article

1) 2/7 2) 3/7 3) 4/7 4) None of these Solution: Option (1) 2/7 Total cases = 7P5, where P denotes permutation. The number of ways... View Article

1) 0.55 2) 0.44 3) 0.22 4) 0.33 Solution: Option (1) 0.55 Total cases = 100C2 Favourable cases = 33C1.67C1 +33C2 (We know... View Article

1) 1/729 2) 1/24 3) 1/21 4) 2/7 Solution: Option (4) 2/7 Total cases = 9C3 Favourable cases = 2C1 . 3C1. 4C1 Hence, the... View Article

1) 1/17 2) 3/19 3) 2/19 4) 2/17 Solution: Option (4) 2/17 Total number of cards = 52 Total cases = 52C3 Total number of red... View Article

1) 24/144 2) 56/144 3) 68/144 4) 76/144 Solution: Option (4) 76/144 The probability that both are apples = (5/12) . (4/12) = 20/144... View Article

1) 3/16 2) 3/5 3) 1/60 4) 1/56 Solution: Option (4) 1/56 Total number of balls = 16 balls Total cases = 16C3 Favourable cases =... View Article

1) 1/7 2) 2/7 3) 3/7 4) 4/7 Solution: Option (1) 1/7 Total number of balls = 4+3 = 7 Hence, the probability that both the balls are... View Article