ProbabilityPage 1
1.  In a simultaneous throw of two coins, the probability of getting at least one tail is:  (a)1/3  (b)1/2  (c)2/3  (d)3/4 

Answer is: DHere S = {HH, HT, TH, TT}
Let E = event of getting at least one tail = {HT, TH, TT}
∴ P(E) = n(E)/n(S) = 3/4
2.  Three unbiased coins are tossed. What is the probability of getting at least 2 heads?  (a)1/4  (b)1/2  (c)1/3  (d)1/8 

Answer is: BHere S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}
∴ P(E) = n(E)/n(S) = 4/8 = 1/2.
3.  What is the probability of getting a sum 9 from two throws of a dice?  (a)1/6  (b)1/8  (c)1/9  (d)1/12 

Answer is: CIn two throws of a die, n(S) = (6 x 6) = 36
Let E = event of getting a sum 9 = {(3, 6), (4, 5), (5, 4), (6, 3)}
∴ P(E) = n(E)/n(S) = 4/36 = 1/9.
4.  From a pack of 52 cards, two cards are drawn together at random. what is the probability of both the cards being aces?  (a)1/15  (b)25/57  (c)35/256  (d)1/221 

Answer is: DLet S be the sample space. Then,
n(S) = 52C2 = (52 x 51)/(2 x 1) = 1326
Let E = event of getting 2 aces out of 4.
∴ n(E) = 4C2 = (4 x 3)/(2 x 1) = 6
∴ P(E) = n(E)/n(S) = 6/1326 = 1/221.
5.  Two cards are drawn together from a pack of 52 cards. The probability that one is a diamond and one is a heart, is:  (a)3/20  (b)29/34  (c)47/100  (d)13/102 

Answer is: DLet S be the sample space. Then,
n(S) = ^{52}C_{2} = (52 x 51)/(2 x 1) = 1326
Let E = event of getting 1 diamond and 1 heart.
∴ n(E) = number of ways of choosing 1 diamond out of 13 and 1 heart out of 13 = (^{13}C_{1} x ^{13}C_{1}) = (13 x 13) = 169.
∴ P(E) = n(E)/n(S) = 169/1326 = 13/102.
6.  A bag contains 5 red, 4 blue and 3 black marbles. Three marbles are drawn at random. What is the probability that they are not of the same colour?  (a)3/44  (b)3/55  (c)52/55  (d)41/44 

Answer is: DLet S be the sample space. Then,
n(S) = number of ways of drawing 3 marbles out of 12 ^{12}C_{3}
n(S) = (12 x 11 x 10)/(3 x 2 x 1) = 220
Let E be the event of drawing 3 marbles of the same color.
Then, E = event of drawing (3 balls out of 5) or (3 balls out of 4) or (3 balls out of 3)
⇒ n(E) = (^{5}C_{3} +^{ 4}C_{3} + ^{3}C_{3}) = (^{5}C_{2} + ^{4}C_{1} + 1) = [(5 x 4)/(2 x 1)] + 4 + 1 = 15
⇒ P(E) = n(E)/n(S) = 15/220 = 3/44.
∴ Required probability = (1 – 3/44) = 41/44.
7.  A box contains 4 green, 5 white and 6 black balls. Three balls are drawn at random from the box. The probability that all of them are white is:  (a)1/22  (b)3/22  (c)2/91  (d)2/77 

Answer is: CLet S be the sample space. Then,
n(S) = number of ways of drawing 3 balls out of 15 = ^{15}C_{3} = (15 x 14 x 13)/(3 x 2 x 1) = 455.
Let E = event of getting all the 3 white balls.
∴ n(E) = ^{5}C_{3} = ^{5}C_{2} = (5 x 4)/(2 x 1) = 10.
∴ P(E) = n(E)/n(S) = 10/455 = 2/91.
8.  A bag contains 2 yellow, 3 green and 2 red balls. Two balls are drawn at random. What is the probability that none of the balls drawn is red?  (a)10/21  (b)11/21  (c)2/7  (d)5/7 

Answer is: ATotal number of balls = (2 + 3 + 2) = 7
Let S be the sample space. Then,
n(S) = number of ways of drawing 2 balls out of 7 = ^{7}C_{2} = (7 x 6)/(2 x 1) = 21.
Let E = event of drawing 2 balls, none of which is red.
∴ n(E) = number of ways of drawing 2 balls out of (2 + 3) balls
= ^{5}C_{2} = (5 x 4)/(2 x 1) = 10
∴ P(E) = n(E)/n(S) = 10/21.