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Practice Questions

1.

In a simultaneous throw of two coins, the probability of getting at least one tail is:
A.1/3B.1/2
C.2/3D.3/4

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Answer is: D

Here 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.1/4B.1/2
C.1/3D.1/8

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Answer is: B

Here 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.1/6B.1/8
C.1/9D.1/12

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Answer is: C

In 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.1/15B.25/57
C.35/256D.1/221

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Answer is: D

Let 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.3/20B.29/34
C.47/100D.13/102

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Answer is: D

Let S be the sample space. Then,
n(S) = 52C2 = (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 = (13C1 x 13C1) = (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.3/44B.3/55
C.52/55D.41/44

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Answer is: D

Let S be the sample space. Then,
n(S) = number of ways of drawing 3 marbles out of 12 12C3
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) = (5C3 + 4C3 + 3C3) = (5C2 + 4C1 + 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.1/22B.3/22
C.2/91D.2/77

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Answer is: C

Let S be the sample space. Then,
n(S) = number of ways of drawing 3 balls out of 15 = 15C3 = (15 x 14 x 13)/(3 x 2 x 1) = 455.
Let E = event of getting all the 3 white balls.
∴ n(E) = 5C3 = 5C2 = (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.10/21B.11/21
C.2/7D.5/7

Report! Answer!

Answer is: A

Total 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 = 7C2 = (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
= 5C2 = (5 x 4)/(2 x 1) = 10
∴ P(E) = n(E)/n(S) = 10/21.

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