13
Q:

An urn contains 4 white 6 black and 8 red balls . If 3 balls are drawn one by one without replacement, find the probability of getting all white balls.

A) 5/204 B) 1/204
C) 13/204 D) None of these

Answer:   B) 1/204



Explanation:

Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then 

Required probability = 

Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9

When  a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white

Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.

Hence the required probability = 

Q:

A room contains 3 brown, 5 black and 4 white chairs. Two chairs are picked and are put in the lawn. What is the probability that none of the chairs picked is white ?

A) 14/33 B) 14/55
C) 12/55 D) 13/33
 
Answer & Explanation Answer: A) 14/33

Explanation:

Total number of chairs = (3 + 5 + 4) = 12.

Let S be the sample space.

Then, n(s)= Number of ways of picking 2 chairs out of 12

12×11/2×66

Let n(E) = number of events of selecting 2 chairs for selecting no white chairs.

=> 8C8×7/2×28

Therefore required probability = 28/66 = 14/33.

Report Error

View Answer Workspace Report Error Discuss

5 224
Q:

The probability that a bowler bowled a ball from a point will hit by the batsman is ¼. Three such balls are bowled simultaneously towards the batsman from that very point. What is the probability that the batsman will hit the ball ?

A) 37/64 B) 27/56
C) 11/13 D) 9/8
 
Answer & Explanation Answer: A) 37/64

Explanation:

Probability of not hitting the ball = 1- 1/4 =IBPS RRB Clerk Level Quiz : Quantitative Aptitude | 11 -09 - 17
Then, the probability that the batsman will hit the ball =IBPS RRB Clerk Level Quiz : Quantitative Aptitude | 11 -09 - 17

Report Error

View Answer Workspace Report Error Discuss

6 243
Q:

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn find the probability the card drawn is red.

A) 2/3 B) 1/2
C) 1/52 D) 13/51
 
Answer & Explanation Answer: B) 1/2

Explanation:

We know that:

When one card is drawn from a pack of 52 cards

The numbers of possible outcomes n(s) = 52

We know that there are 26 red cards in the pack of 52 cards
⇒ The numbers of favorable outcomes n(E) = 26

Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S)

∴ required probability = 26/52 = 1/2.

Report Error

View Answer Workspace Report Error Discuss

5 334
Q:

A fair coin is tossed 11 times. What is the probability that only the first two tosses will yield heads ?

A) (1/2)^11 B) (9)(1/2)
C) (11C2)(1/2)^9 D) (1/2)
 
Answer & Explanation Answer: A) (1/2)^11

Explanation:

Probability of occurrence of an event,

P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S)


⇒ Probability of getting head in one coin = ½,

⇒ Probability of not getting head in one coin = 1- ½ = ½,

Hence,

All the 11 tosses are independent of each other.

∴ Required probability of getting only 2 times heads =

Report Error

View Answer Workspace Report Error Discuss

5 291
Q:

The first 8 alphabets are written down at random. what is the probability that the letters b,c,d,e always come together ?

A) 1/7 B) 8!
C) 7! D) 1/14
 
Answer & Explanation Answer: D) 1/14

Explanation:

The 8 letters can be written in 8! ways.
n(S) = 8!
Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.
Now the letters a, (bcde), f, g and h can be arranged in 5! ways.
The letters b,c,d and e can be arranged themselves in 4! ways.
n(E) = 5! x 4!
Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14
Hence the answer is 1/14. 

Report Error

View Answer Workspace Report Error Discuss

13 731