# Probability

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3736 Directions: Each of the questions given below is followed by series of options among which you have to choose the best one. You can also review your choices by clicking into answer section.

1. A card is drawn at random from a pack of well-shuffled cards. What is the probability that the card is a face card?

A. 2/13

B. 1/13

C. 4/13

D. 3/13

Ans: Option D

Explanation:

There are 3 face cards in each suit. Hence, a total of 12 face cards in a pack. P(E) = 12/52 = 3/13

2. 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. 1/3

B. 3/4

C. 7/19

D. 8/21

E. 9/21

Ans: Option A

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 blue.

n(E) = 7.

P(E) =   n(E)/ n(S)= 7/21 =1/3

3. What is the probability of selecting a prime number from 1, 2, 3 ,… 10 ?

A. 2/5

B. 1/5

C. 3/5

D. 1/7

Ans: Option A

Explanation:

Total count of numbers, n(S) = 10

Prime numbers in the given range are 2 , 3, 5 and 7

Hence, total count of prime numbers in the given range, n(E) = 4

P(E) = n(E)/n(S)=4/10=2/5

4. One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?

A. 1/13

B. 3/13

C. 1/4

D. 9/52

Ans: Option B

Explanation:

Clearly, there are 52 cards, out of which there are 12 face cards.

P(getting a face card). =12/52 = 3/13

5. John and Dani go for an interview for two vacancies. The probability for the selection of John is 1/3 and whereas the probability for the selection of Dani is 1/5. What is the probability that none of them are selected?

A. 3/5

B. 7/12

C. 8/15

D. 1/5

Ans: Option C

Explanation:

Let A = the event that John is selected and B = the event that Dani is selected.

Given that P(A) = 1/3 and P(B) = 1/5

We know that A' is the event that A does not occur and B' is the event that B does not occur

Probability that none of them are selected

=P(A'∩B')(∵ Reference : Algebra of Events)

=P(A').P(B') (∵ Here A and B are Independent Events and refer theorem on independent events)

=[ 1 – P(A)][ 1 – P(B)]

=1−(1/3) x 1−(1/5)

=(2/3)×(4/5)=8/15

6. A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective is

A. 4/19

B. 7/19

C. 12/19

D. 21/95

Ans: Option B

Explanation:

P(No bulb is defective) = 16C2/20C2 = 12/19

P(At least one is defective) = 1-(12/19) = 7/19

7. 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

Ans: Option C

Explanation:

Let S be the sample space.

Then, n(S)= number of ways of drawing 3 balls out of 15

= 15C3

=(15 x 14 x 13)/(3 x 2 x 1)

= 455.

Let E = event of getting all the 3 red balls.

n(E) = 5C3 = 5C2 = (5 x 4)/ (2 x 1)= 10.

P(E) =    n(E)/ n(S)=10/455=2/91

8. There are 4 hotels in a town. If 3 men check into the hotels in a day, find the probability that each checks into a different hotel.

A. 3/4

B. 5/8

C. 3/8

D. 1/4

Ans: Option C

Explanation:

No. of ways in which 3 men can check into 4 hotels n(S) = 4*4*4 = 64

No. of ways in which they check into different hotels n(E) = 4*3*2 = 24 P(E) = 24/64 = 3/8

9. In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

A.1/10

B.2/5

C.2/7

D.5/7

Ans: Option C

Explanation:

P (getting a prize) =10/(10 + 25)   =10/35  =2/7

10. A letter is chosen at random from the word 'ASSASSINATION'. What is the probability that it is a vowel?

A. 4/13

B. 8/13

C. 7/13

D. 6/13

Ans: Option D

Explanation:

Total Number of letters in the word ASSASSINATION, n(S) = 13

Total number of Vowels in the word ASSASSINATION, n(E) = 6 (∵ 3 'A', 2 'I', 1 'O')

Probability for getting a vowel, P(E) =n(E)/n(S)=6/13

11. Two dice are tossed. The probability that the total score is a prime number is:

A. 1/6

B. 5/12

C. 1/2

D. 7/9

Ans: Option B

Explanation:

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

Then E   = { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3),  (5, 2), (5, 6), (6, 1), (6, 5) }

n(E) = 15.

P(E) =   n(E)/ n(S)=15/36=5/12

.12. When two dice are tossed, what is the probability that the total score is a prime number?

A. 14

B. 13

C. 23

D. 512

Ans: Option D

Explanation:

Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces)

Hence, Total number of outcomes possible when two dice are rolled, n(S) = 6 × 6 = 36

Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 … etc

Let E = the event that the total is a prime number

= {(1,1), (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4),

(4,1), (4,3), (5,2), (5,6), (6,1), (6,5)}

Hence, n(E) = 15

P(E) = n(E)/n(S)=15/36=5/12

13. Two brothers appeared for an exam. The probability of each of them getting selected is 1/4 and 1/6 respectively. Find the probability that only one of the two are selected.

A. 1/24

B. 1/12

C. 10/24

D. 1/3

Ans: Option D

Explanation:

Probability that only of the two are selected = P(A alone selected).+P(B alone selected)

= P(A selected)*P(B not selected)+ P(A not selected).*P(B selected).

= [(1/4)*(5/6)]+[(3/4)*(1/6)] = [5/24]+[3/24] = 8/24 = 1/3

14. Six persons enter a lift on the ground floor of a nine floor apartment. Assuming that each of them independently and with equal probability can leave the lift at any floor beginning with the first, what is the probability that all the six persons are leaving the lift at different floors?

A. 8!86

B. 8!87

C. 8C686

D. 8P686

Ans: Option D

Explanation:

Apart from the ground floor, there are 8 floors

Let's find out the total number of ways in which all the six persons can leave the lift at eight different floors

The 1st person can leave the lift in any of the 8 floors (8 ways)

The 2nd person can leave the lift in any of the remaining 7 floors (7 ways)

The 3nd person can leave the lift in any of the remaining 6 floors (6 ways)

…

The 6th person can leave the lift in any of the remaining 3 floors (3 ways)

Total number of ways = 8 × 7 × 6 × 5 × 4 × 3 = 8P6

(In fact, from the definition of permutations itself, we will be able to directly say that the number of ways in which all the six persons can leave the lift at 8 different floors = 8P6)

n(E)= Total Number of ways in which all the six persons can leave the lift at eight different floors = 8P6

Now we will find out the total number of ways in which each of the six persons can leave the lift at any of the eight floors

The 1st person can leave the lift in any of the 8 floors (8 ways)

The 2nd person can leave the lift in any of the 8 floors (8 ways)

The 3nd person can leave the lift in any of the 8 floors (8 ways)

…

The 6th person can leave the lift in any of the 8 floors (8 ways)

Total number of ways = 8 × 8 × 8 × 8 × 8 × 8 = 86

i.e., The total number of ways in which each the six persons can leave the lift

at any of the eight floors = n(S) = 86

P(E) = n(E)n(S)=8P686

15. A bag contains 4 pink and 5 blue beads. Another bag contains 3 pink and 6 blue beads. Two beads are drawn, one from each bag. Find the probability that one is pink and the other is blue.

A. 13/27

B. 1/2

C. 7/11

D. 14/27

Ans: Option A

Explanation:

n(S) = 9C1*9C1 = 81

No. of ways of selecting a pink and a blue bead = [n(Pink from bag 1)*n(Blue from bag 2)]+ [n(Blue from bag 1)*n(Pink from bag 2)]

n(E) = (4*6)+(5*3) = 39

P(E) = n(E)/ n(S) = 13/27

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