Probability Quiz for RBI Assistant Mains / IBPS SO 2016-17

1. A bag contains 5 red and 7 white balls. Four balls are drawn out one by one and not replaced. What is the probability that they are alternatively of different colours?

a) 7/99
b) 11/99
c) 14/99
d) 19/99
e) None of these

2. P and Q are sitting in a ring with 11 other persons. If the arrangement of 11 persons is at random, then the probability that there are exactly 4 persons between them?

a) 1/3
b) 1/4
c) 1/5
d) 1/6
e) None of these

3. 10 persons are seated around a round table. What is the probability that 4 particular persons are always seated together?

a) 1/21
b) 4/21
c) 8/21
d) 11/21
e) None of these

4. A box contains 4 red, 5 black and 6 green balls. 3 balls are drawn at random. What is the probability that all the balls are of same colour?

a) 33/455
b) 34/455
c) 44/455
d) 47/455
e) None of these

5. An apartment has 8 floors. An elevator starts with 4 passengers and stops at 8 floors of the apartment. What is the probability that all passengers travels to different floors?

a) 109/256
b) 135/256
c) 105/256
d) 95/256
e) None of these

6. A speak truth in 60% cases and B in 80% cases. In what percent of cases they likely to contradict each other narrating the same incident?

a) 9/25
b) 7/25
c) 11/25
d) 13/25
e) None of these

7. A box contains 30 electric bulbs, out of which 8 are defective. Four bulbs are chosen at random from this box. Find the probability that at least one of them is defective?

a) 432/783
b) 574/783
c) 209/784
d) 334/784
e) None of these

8. Two person A and B appear in an interview.  The probability of A’s selection is 1/5 and the probability of B’s selection is 2/7. What is the probability that only one of them is selected?

a) 11/35
b) 12/35
c) 13/35
d) 17/35
e) None of these

9. A 4- digit number is formed by the digits 0, 1, 2, 5 and 8 without repetition. Find the probability that the number is divisible by 5.

a) 1/5
b) 2/5
c) 3/5
d) 4/5
e) None of these

10. A bag contains 6 red balls and 8 green balls. 2 balls are drawn at random one by one with replacement. Find the probability that both the balls are green

a) 16/49
b) 25/49
c) 12/49
d) 21/49
e) None of these


ANSWERS :

1. Balls are picked in two manners – RWRW or WRWR
So probability = (5/12)*(7/11)*(4/10)*(6/9) + (7/12)*(5/11)*(6/10)*(4/9) = 14/99.

2. Fix the position of P, then Q can be sit in 12 positions, so total possible outcome = 12
Now, exactly 4 persons are sitting between them. This can be done in two ways as shown in figure, so favourable outcomes = 2
So, probability = 2/12 = 1/6.

3. Total outcomes = (10 -1)! = 9!
Favourable outcomes = 6!*4! (4 person seated together and 6 other persons seated randomly, so they will sit in (7-1)! Ways and those 4 persons can be arranged in 4! ways)
So probability = 1/21.

4. (4c3 + 5c3 + 6c3)/15c3 = 34/455.

5. Total outcomes = 8*8*8*8
Favourable outcomes = 8*7*6*5 (first person having 8 choices, after that second person have 7 choices and so on)
So, probability = 105/256.

6. P(A) = 3/5 and P(B) = 4/5. Now they are contradicting means one is telling truth and other telling the lie. So,
Probability = (3/5)*(1/5) + (2/5)*(4/5) = 11/25

7. 1 – 22c4/30c4 = 1 – 209/783 = 574/783.

8. A selects and B rejects + B selects and A rejects = (1/5)*(5/7) + (4/5)*(2/7) = 13/35.

9. Total possibility = 5*4*3*2
Favourable outcomes = 2*4*3*2  (to be divisible by 5 unit digit can be filled with only 0 or 5, so only two possibilities are there,  then the remaining can be filled in 4, 3 and 2 ways respectively)
so probability = 2/5.

10. (8c1)/(14c1) * (8c1)*(14c1) = 16/49.

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