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1) Bag A contains 5 blue and 6 red balls, bag B contains 7 blue and 5 red balls. If 6 balls are drawn at random from each of the bags, find the product of the probability that 4 blue and 2 red balls are drawn from bag A and 3 blue and 3 red balls are drawn from bag B.
A.1621/35802
B.1750/35574
C.1623/35082
D.1619/35820
E.none of these
2) A bag contains 6 apple, 5 Banana and (x+4) Orange. If two fruits are taken at random and the probability of getting both are apple is 1/14, then find the value of x.
A.5
B.6
C.4
D.8
E.None of these
3) The box contains a certain number of red balls and certain number of green balls. If two balls are selected randomly, the probability that balls being red is 1/6, then find the total number of balls in that box?
A.10
B.8
C.7
D.5
E.Cannot be determined
4) If 4 cards are drawn at random from a pack of cards, find the probability that at least 3 cards are spade cards.
A.913/15000
B.813/17000
C.713/17000
D.913/16000
E.913/20825
5) A bag contains 6 red balls, three green balls and three pink balls. If two balls are drawn at random, then what is the probability that one ball is pink and other is green or one ball is red and other is pink?
A.9/22
B.8/23
C.10/23
D.9/19
E.None of these
6) A bag contains 4 red balls and 4 yellow balls. If four balls are picked at random, then what is the probability that balls picked alternatively are of different colour, if repetition not allowed?
A.1/7
B.6/35
C.1/5
D.8/35
E.None of these
7) An urn consists of some green, red and black balls. Total number of balls in the urn is 15. If two balls are picked, Probability that the two balls being green is 2/21. Now one ball is picked from the urn, find the probability that the ball is either red or black.
A.Cannot be determined
B.2/15
C.1/5
D.1/10
E.2/3
8) Two persons stole the gold chain. The probability of catching the first person is ¼. The probability of catching the second person is 1 / 6. What is the probability that only one of them is caught?
A.6 / 12
B.2 / 3
C.1 / 3
D.5 / 12
E.None of these
9) A bag contains (x + 3) pink, 4 blue and 6 white color balls. If two balls are taken random and the probability of getting both being blue color balls is 2/51, then find the total number of pink color balls?Â
A.6 balls
B.7 balls
C.8 balls
D.5 balls
E.None of these
10) There are three groups of fire extinguishers in a city i.e. P, Q and R. In group P there are (x-6) females and (x+21) males. In group Q, there are 3 females and 12 males whereas in group R, there are 5 females and 7 males. One person is selected at random from each group. Find the number of females in group P, if the probability of selecting all three males is 4/15.
A.81
B.85
C.83
D.71
E.89
Answers :
1) Answer: B
Probability of drawing 4 blue and 2 red balls from bag A = (5c4 x  6c2)/11c6 = 25/154
Probability of drawing 3 blue and 3 red balls from bag B = (7c3 x  5c3)/12c6 = 25/66
Required product = 25/154 x 25/66 = 625/10164.
2) Answer: B
6C2/(15 + x)C2Â = 1/14
30 * 14 = (15 + x) (14 + x)
30 * 14 = 210 + 15x + 14x + x2
x2 + 29x – 210 = 0
x2 + 35x – 6x – 210 = 0
x(x + 35) – 6(x + 35) = 0
(x – 6)(x + 35) = 0
x = 6, -35(negative value neglected)
3) Answer: E
Number of red balls = x
Number of green balls = y
xC2/(x + y) C2Â = 1/6
Data inadequate
4) Answer: E
Possible cases = 3 spade cards + any other card OR 4 spade cards
Required probability = (13C3*39C1)/52C4 + (13C4/52C4)
Required probability = {(286*39)/52C4} + (715/52C4)
Required probability = {11869/52C4}
Required probability = {11869/(52*51*50*49/4*3*2)}
Required probability = {11869/13*17*25*49}
Required probability = {913/17*25*49}
Required probability = {913/20825}
5) Answer: A
Required probability = (3C1Â * 3C1Â + 6C1Â * 3C1)/12C2
= 9/22
6) Answer: B
Required probability = (4C1/8C1Â * 4C1/7C1Â * 3C1/6C1Â * 3C1/5C1) + (4C1/8C1Â * 4C1/7C1Â * 3C1/6C1Â * 3C1/5C1)
= ½ * 4/7 * ½ * 3/5 + ½ * 4/7 * ½ * 3/5
= 12/140 + 12/140
= 6/35
7) Answer: E
Let G, R and B be the number of green, red and black balls respectively.
G + R + B = 15
P (two balls are green) = 2/21
G*(G-1)/(15*14) = 2/21
G2 – G = 20
G2 – G – 20 = 0
G2 – 5G + 4G – 20 = 0
G(G-5) + 4(G-5) = 0
(G+4)*(G-5) = 0
G = 5
R + B = 10
P (Either red or black) = 10/15 = 2/3
8) Answer: C
P(1st not caught) = 1 – 1 / 4 = 3 / 4
P(2nd not caught) = 1 – 1 / 6 = 5 / 6
P(Only one caught) = P(1st caught and 2nd not caught) + P(2nd caught and 1st not caught)
(1 / 4 * 5 / 6) + (3 / 4 * 1 / 6) = 1 / 3
9) Answer: C
4C2Â / (x + 13)C2Â = 2/51
[(4 * 3) / (1 * 2)] / [(x + 13) (x + 12) / (1 * 2)] = 2/51
(51 * 6) = x2Â + 13x + 12x + 156
306 = x2Â + 25x + 156
x2 + 25x – 150 = 0
(x – 5) (x + 30) = 0
x = 5, -30 (negative value will be eliminated)
The total number of pink color balls = 8 balls
10) Answer: A
{(x + 21)/ (x – 6 + x + 21)} * (12/15) * (7/12) = 4/15
{(x+21)/ (2x+15)}*(7/15) = 4/15
{(x+21)/ (2x+15)} = 4/7
7x+147 = 8x+60
x = 87
Number of females in group P = 87 – 6 = 81
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