Probability ( 16 to 20 )

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Probability ( 16 to 20 ):


APTITUDE QUESTIONS (PROBABILITY):

16. The probability that a card drawn from a pack of 52 cards will be a diamond or a king is:

A)    (2/13)

B)    (4/13)

C)    (1/13)

D)    (4/52)

E)    None of these

View Answer

Answer: option (B)

Explanation:

Here, n(S) = 52

There are 13 cards of diamond (including one king) and then there are more kings.

Let E = event of getting a diamond or a king.

Then, n(E) = (13+3) = 16

∴ P(E) = (n(E)/n(S)) = (16/52) = (4/13)

 

17. From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

A)    (1/15)

B)    (25/57)

C)    (35/256)

D)    (1/221)

E)    None of these

View Answer

Answer: option (D)

Explanation:

Let S be the sample space. Then,

n(S) = 52C2 = ((52*51)/(2*1)) = 1326

Let E = event of getting 1 spade and 1 heart.

∴ n(E) = 4C2 = ((4*3)/(2*1)) = 6

∴ P(E) = (n(E)/n(S)) = (6/1326) = (1/221)

 

18. A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

A)    (3/4)

B)    (4/7)

C)    (1/8)

D)    (3/7)

E)    None of these


View Answer

Answer: option (B)

Explanation:

Total number of balls = (6 + 8) = 14

Number of white balls = 8

∴ P(drawing a white ball) = (8/14) = (4/7)

 

19. In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

A)    (2/3)

B)    (3/4)

C)    (7/19)

D)    (8/21)

E)    None of these

View Answer

Answer: option (D)

Explanation:

Total number of balls = (8 + 7 + 6) = 21

Let E = event that the ball drawn is neither red nor green

= event that the ball drawn is red

∴ n(E) = 8

∴ P(E) = (8/21)

 

20. Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:

A)    (3/20)

B)    (29/34)

C)    (47/100)

D)    (13/102)

E)    None of these

View Answer

Answer: option (D)

Explanation:

Let S be the sample space. Then,

n(S) = 52C2 = ((52*51)/(2*1)) = 1326

Let E = event of getting 1 spade and 1 heart.

∴ n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13

= (13C1 * 13C1) = (13 * 13) = 169

∴ P(E) = (n(E)/n(S)) = (169/1326) = (13/102)

 

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