Aptitude - Probability - Discussion
Discussion Forum : Probability - General Questions (Q.No. 1)
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?
Answer: Option
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}.
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n(E) | = | 9 | . |
n(S) | 20 |
Discussion:
101 comments Page 1 of 11.
PURVA said:
2 months ago
@All.
Here is my solution.
P(A or B) = P(A) + P(B) -P(A AND B).
P(3) = 6/20 {3,6,9,12,15,18},
P(5) = 4/20 {5,10,15,20},
P(3 and 5) = 1/20 {15}.
Here is my solution.
P(A or B) = P(A) + P(B) -P(A AND B).
P(3) = 6/20 {3,6,9,12,15,18},
P(5) = 4/20 {5,10,15,20},
P(3 and 5) = 1/20 {15}.
(4)
Sriram said:
3 months ago
The multiple of 3 and 5 is given numbers 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,7,18,19,20. While the question is about the multiple of 3 is 3,6,9,12,15,18 and the multiple of 5 is 5,10,15,20.
The total number is 10.
So,
10/20 = 1/2.
The total number is 10.
So,
10/20 = 1/2.
(5)
Abhishek tomar said:
10 months ago
15 is the multiple of 3 and 5. while question is about the multiple of 3 or 5 so, why not option C?
(3)
Niraml said:
11 months ago
Here, the total ticket is 20 we gonna take 1 from that 20 tickets it can be multiple of 3 or 5
so,
5C1 + 3C1/20C1 = 5+3/20 = 8/20.
= 2/5 is correct.
so,
5C1 + 3C1/20C1 = 5+3/20 = 8/20.
= 2/5 is correct.
Avaneet Agrahari said:
3 years ago
Multiple of 3=[,3,6,912,15,18]=6
M....o. 5=[5,10,15,20]=4 but 15 is already in 3 so it count 3
Ans= 6+3/20=9/20
M....o. 5=[5,10,15,20]=4 but 15 is already in 3 so it count 3
Ans= 6+3/20=9/20
(3)
Kavya said:
3 years ago
@Uday,
15 is repeated.
And 5 is missing.
Anyway, the answer is correct.
15 is repeated.
And 5 is missing.
Anyway, the answer is correct.
(3)
Kousalya said:
3 years ago
Yes, I too agree 1/2 is the right answer.
Uday said:
4 years ago
(3, 6,9,12,15,18,5,10,15,20).
= 10/20,
= 1/2.
= 10/20,
= 1/2.
Shivu said:
4 years ago
Ans is 9/20 because;
Multiple of 3 (3 6 9 12 15 18)=possiblity 6 so 6/20,
Multiple of 5 (5 10 15 20)=possibility 4 so 4/20,
15 is both sides so eliminate, so -1/20.
6/20+4/20-1/20 = 9/20.
Multiple of 3 (3 6 9 12 15 18)=possiblity 6 so 6/20,
Multiple of 5 (5 10 15 20)=possibility 4 so 4/20,
15 is both sides so eliminate, so -1/20.
6/20+4/20-1/20 = 9/20.
(2)
Anitha said:
4 years ago
@Shraddha.
5 is also considered so that we get probability 9/20. If 5 is not considered we can't get 9/20.
5 is also considered so that we get probability 9/20. If 5 is not considered we can't get 9/20.
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