Probability ( 21 to 25 )

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Probability ( 21 to 25 ):


APTITUDE QUESTIONS (PROBABILITY):

21. A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

A)    (1/22)

B)    (3/22)

C)    (2/91)

D)    (2/77)

E)    None of these

View Answer

Answer: option (C)

Explanation:

Let S be the sample space. Then,

n(S) = number f ways of drawing 3 balls out of 15 = 15C3 = ((15*14*13)/(3*2*1)) = 455

Let E = event of getting all the 3 red balls

∴ n(E) = 5C3 = 5C2 = ((5*4)/(2*1)) = 10

∴ P(E) = (n(E)/n(S)) = (10/455) = (2/91)

 

22. A bag contains 6 white and 4 red balls. Three red balls are drawn at random. What is the probability that one ball is red and the other two are white?

A)    (1/2)

B)    (1/12)

C)    (3/10)

D)    (7/12)

E)    None of these

View Answer

Answer: option (A)

Explanation:

Let S be the sample space. Then,

n(S) = number f ways of drawing 3 balls out of 10 = 10C3 = ((10*9*8)/(3*2*1)) = 120

Let E = event of drawing 1 red ball and 2 white balls

∴ n(E) = number of ways of drawing 1 red ball out of 4 and 2 white balls out of 6

= (4C1 * 6C2) = (4*(6*5/2*1)) = 60

∴ P(E) = (n(E)/n(S)) = (60/120) = (1/2)

 

23. A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

A)    (10/21)

B)    (11/21)

C)    (2/7)

D)    (5/7)

E)    None of these

View Answer

Answer: option (A)


Explanation:

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*6)/(2*1)) = 21

Let E = event of drawing 2 balls, none of which is blue.

∴ n(E) = number of ways of drawing 2 balls out of (2 + 3) balls

= 5C2 = ((5*4/2*1)) = 10

∴ P(E) = (n(E)/n(S)) = (10/21)

 

24. A box contains 10 black and 10 white balls. The probability of drawing two balls of the same color, is:

A)    (9/19)

B)    (9/38)

C)    (10/19)

D)    (5/19)

E)    None of these

View Answer

Answer: option (A)

Explanation:

Total number of balls = 20

Let S be the sample space. Then,

n(S) = number of ways of drawing 2 balls out of 20 = 20C2 = ((20*19)/(2*1)) = 190

Let E = event of drawing 2 balls of the same color

∴ n(E) = (10C2 + 10C2) = 2 * ((10*9/2*1)) = 90

∴ P(E) = (n(E)/n(S)) = (90/190) = (9/19)

 

25. In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 bots are selected is:

A)    (21/46)

B)    (25/117)

C)    (1/50)

D)    (3/25)

E)    None of these

View Answer

Answer: option (A)

Explanation:

Let S be the sample space. Then,

n(S) = number of ways of selecting 3 students out of 25 = 25C3 = ((25*24*23)/(3*2*1)) = 2300

Let E = event of selecting 1 girl and 2 boys.

∴ n(E) = (10C1 * 15C2) = 10 * ((15*14/2*1)) = 1050

∴ P(E) = (n(E)/n(S)) = (1050/2300) = (21/46)

 

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