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- Question
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?
Options- A.
1 2 - B.
2 5 - C.
8 15 - D.
9 20 - Correct Answer
-
9 20
ExplanationHere, 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
Probability problems
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- 1. 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:
Options- A.
1 22 - B.
3 22 - C.
2 91 - D.
2 77
Discuss
Correct Answer:2 91
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 - 2. In 100 m race, A covers the distance in 36 seconds and B in 45 seconds. In this race A beats B by:
Options- A. 20 m
- B. 25 m
- C. 22.5 m
- D. 9 m Discuss
Correct Answer: 20 m
Explanation:
Distance covered by B in 9 sec. = ❨ 100 x 9 ❩m = 20 m. 45 ∴ A beats B by 20 metres.
- 3. In a game of 100 points, A can give B 20 points and C 28 points. Then, B can give C:
Options- A. 8 points
- B. 10 points
- C. 14 points
- D. 40 points Discuss
Correct Answer: 10 points
Explanation:
A : B = 100 : 80.A : C = 100 : 72.
∴ B = ❨ B x A ❩ = ❨ 80 x 100 ❩ = 10 = 100 = 100 : 90. C A C 100 72 9 90 ∴ B can give C 10 points.
- 4. A runs 1⅔ times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?
Options- A. 200 m
- B. 300 m
- C. 270 m
- D. 160 m Discuss
Correct Answer: 200 m
Explanation:
Ratio of the speeds of A and B = 5 : 1 = 5 : 3. 3 Thus, in race of 5 m, A gains 2 m over B.
2 m are gained by A in a race of 5 m.
80 m will be gained by A in race of ❨ 5 x 80 ❩m = 200 m. 2 ∴ Winning post is 200 m away from the starting point.
- 5. In a 100 m race, A beats B by 10 m and C by 13 m. In a race of 180 m, B will beat C by:
Options- A. 5.4 m
- B. 4.5 m
- C. 5 m
- D. 6 m Discuss
Correct Answer: 6 m
Explanation:
A : B = 100 : 90.A : C = 100 : 87.
B = B x A = 90 x 100 = 30 . C A C 100 87 29 When B runs 30 m, C runs 29 m.
When B runs 180 m, C runs ❨ 29 x 180 ❩m = 174 m. 30 ∴ B beats C by (180 - 174) m = 6 m.
- 6. Two dice are tossed. The probability that the total score is a prime number is:
Options- A.
1 6 - B.
5 12 - C.
1 2 - D.
7 9
Discuss
Correct Answer:5 12
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 - 7. 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)?
Options- A.
1 13 - B.
3 13 - C.
1 4 - D.
9 52
Discuss
Correct Answer:3 13
Explanation:
Clearly, there are 52 cards, out of which there are 12 face cards.∴ P (getting a face card) = 12 = 3 . 52 13 - 8. Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:
Options- A.
3 20 - B.
29 34 - C.
47 100 - D.
13 102
Discuss
Correct Answer:13 102
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 - 9. A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:
Options- A.
1 13 - B.
2 13 - C.
1 26 - D.
1 52
Discuss
Correct Answer:1 26
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 - 10. What is the probability of getting a sum 9 from two throws of a dice?
Options- A.
1 6 - B.
1 8 - C.
1 9 - D.
1 12
Discuss
Correct Answer:1 9
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
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