# Probability - MCQ 2

## 20 Questions MCQ Test Quantitative Aptitude for Competitive Examinations | Probability - MCQ 2

Description
This mock test of Probability - MCQ 2 for Quant helps you for every Quant entrance exam. This contains 20 Multiple Choice Questions for Quant Probability - MCQ 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Probability - MCQ 2 quiz give you a good mix of easy questions and tough questions. Quant students definitely take this Probability - MCQ 2 exercise for a better result in the exam. You can find other Probability - MCQ 2 extra questions, long questions & short questions for Quant on EduRev as well by searching above.
QUESTION: 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?

Solution:

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

QUESTION: 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?

Solution:

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

QUESTION: 3

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

Solution:

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

QUESTION: 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?

Solution:

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

QUESTION: 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?

Solution:

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

QUESTION: 6

A speaks truth in 60% of the cases and B in 80% of the cases. In what  percentage of cases are they likely to contradict each other, narrating the same incident?

Solution:

Let  A = Event that A speaks the truth

and B= Event  that B speaks the truth.

Then,  P ( A)  = 60/ 100 = ⅗

P ( B)  = 80/ 100 =  ⅘

_

:-     P  ( A) =  ( 1- ⅗  )  = 2 / 5

_

P ( B )  = (1 - 4 /5 ) =  1 / 5

P ( A and B  contradict each other)

= P [ ( A speaks truth and B tells a lie) or

( A tells the lie and B speaks the truth) ]

_         _

=  P [ ( A and B)  or (  A  and  B)]

_              _

= P ( A and B)  +    P ( A  and B)

_                _

=  P (A) *  P ( B)  +     P ( A) *  P ( B)

=  (  3 / 5 * 1 / 5 ) +      ( 2 / 5 * 4 / 5 )

=  3 / 25 + 8 / 25

=  11 / 25

= [ 11/ 25 * 100 ] %

=  44%

QUESTION: 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?

Solution:

1 – 22c4/30c4 = 1 – 209/783 = 574/783

QUESTION: 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?

Solution:

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

QUESTION: 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.

Solution: QUESTION: 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

Solution:

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

QUESTION: 11

A six-digit is to be formed from the given numbers 1, 2, 3, 4, 5 and 6. Find the probability that the number is divisible by 4.

Solution:

For a number to be divisible by 4, the last two digit should be divisible by 4.
So possible cases – 12, 16, 24, 32, 36, 52, 56, 64 (last two digits)
So favourable outcomes = 24 +24 +24 +24 + 24+ 24+24+24 = 192
So p = 192/720 = 4/15

QUESTION: 12

A bag contains 6 red balls and 7 white balls. Another bag contains 5 red balls and 3 white balls. One ball is selected from each. Find the probability that one ball is red and one is white?

Solution:

(6/13)*(3/8) + (7/13)*(5/8) = 53/104

QUESTION: 13

A lottery is organised by the college ABC through which they will provide scholarship of rupees one lakhs to only one student. There are 100 fourth year students, 150 third year students, 200 second year students and 250 first year students. What is the probability that a second year student is choosen.

Solution:

Second year students = 200
so, P = 200/700 = 2/7

QUESTION: 14

A card is drawn from a pack of 52 cards. The card is drawn at random; find the probability that it is neither club nor queen?

Solution:

1 – [13/52 + 4/52 – 1/52] = 9/13

QUESTION: 15

A box contains 50 balls, numbered from 1 to 50. If three balls are drawn at random with replacement. What is the probability that sum of the numbers are odd?

Solution:

There are 25 odd and 25 even numbers from 1 to 50.
Sum will be odd if = odd + odd + odd, odd + even + even, even + odd + even, even+ even + odd
P = (1/2)*(1/2)*(1/2) + (1/2)*(1/2)*(1/2) + (1/2)*(1/2)*(1/2) + (1/2)*(1/2)*(1/2)
=4/8 = ½

QUESTION: 16

From a pack of cards, if three cards are drawn at random one after the other with replacement, find the probability that one is ace, one is jack and one is queen?

Solution:

(4c1 + 4c1 + 4c1)/(52c3)

QUESTION: 17

A and B are two persons sitting in a circular arrangement with 8 other persons. Find the probability that both A and B sit together.

Solution:

Total outcomes = (10 -1)! = 9!
Favourable outcomes = (9 -1)!*2!
So p = 2/9

QUESTION: 18

Find the probability that in a random arrangement of the letter of words in the word ‘PROBABILITY’ the two I’s come together.

Solution:

Total outcomes = 11!/(2!*2!)
favourable outcomes = (10!*2!)/(2!*2!)
p = 2/11

QUESTION: 19

In a race of 12 cars, the probability that car A will win is 1/5 and of car B is 1/6 and that of car C is 1/3. Find the probability that only one of them won the race.

Solution:

1/5 + 1/6 + 1/3= 7/10 (all events are mutually exclusive)

QUESTION: 20

A bag contains 3 red balls and 8 blacks ball and another bag contains 5 red balls and 7 blacks balls, one ball is drawn at random from either of the bag, find the probability that the ball is red.

Solution:

Probability = probability of selecting the bag and probability of selecting red ball
(1/2)*(3/11) + (1/2)*(5/12) = 91/264