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Instructions: This Monk test contains JK Bank Probability Problems with answer. This will help you in JK bank exam preparation.

This is a FREE online JK Bank Monk Test on topic JK Bank Probability Problems with answer. You DO NOT pay money to anyone to attend this test.
Each question carries 1 mark, no negative marks.
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Directions to Solve
In the following the questions choose the word which best expresses the meaning of the given word.

### #1. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

Explanation:

Here, S = {1, 2, 3, 4, …., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}. P(E) = n(E) = 9 . n(S) 20

### #2. 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?

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 x 6) (2 x 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 x 4) (2 x 1)
= 10. P(E) = n(E) = 10 . n(S) 21

### #3. 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?

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) = 7 = 1 . n(S) 21 3

### #4. What is the probability of getting a sum 9 from two throws of a dice?

Explanation:

In two throws of a dice, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}. P(E) = n(E) = 4 = 1 . n(S) 36 9

### #5. Three unbiased coins are tossed. What is the probability of getting at most two heads?

Explanation:

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}. P(E) = n(E) = 7 . n(S) 8
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### #6. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

Explanation:

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

 Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} n(E) = 27. P(E) = n(E) = 27 = 3 . n(S) 36 4

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

Explanation:

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25
25C3 `
 = (25 x 24 x 23) (3 x 2 x 1)
= 2300.
n(E) = (10C1 x 15C2)
 = 10 x (15 x 14) (2 x 1)
= 1050. P(E) = n(E) = 1050 = 21 . n(S) 2300 46

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

Explanation:

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

### #9. From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

Explanation:

Let S be the sample space.

 Then, n(S) = 52C2 = (52 x 51) = 1326. (2 x 1)

Let E = event of getting 2 kings out of 4. n(E) = 4C2 = (4 x 3) = 6. (2 x 1) P(E) = n(E) = 6 = 1 . n(S) 1326 221

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

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) = 15 = 5 . n(S) 36 12
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### #11. A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:

Explanation:

Here, n(S) = 52.

Let E = event of getting a queen of club or a king of heart.

Then, n(E) = 2. P(E) = n(E) = 2 = 1 . n(S) 52 26

### #12. 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:

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) = 10. (2 x 1) P(E) = n(E) = 10 = 2 . n(S) 455 91

### #13. Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:

Explanation:

Let S be the sample space.

 Then, n(S) = 52C2 = (52 x 51) = 1326. (2 x 1)

Let E = event of getting 1 spade and 1 heart. n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = (13C1 x 13C1) = (13 x 13) = 169. P(E) = n(E) = 169 = 13 . n(S) 1326 102

### #14. 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)?

Explanation:

Clearly, there are 52 cards, out of which there are 12 face cards. P (getting a face card) = 12 = 3 . 52 13