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}.
 P(E) = |
n(E) | = | 9 | . |
| n(S) | 20 |
Discussion:
106 comments Page 1 of 11.
Rajesh Bhadra said:
4 months ago
@Ganapathithanuja.
From 1 to 20 there divisible by 3 is -> 3, 6, 9, 12, 15, 18 and divisible by 5 is -> 5, 10, 15, 20.
Now if you see total numbers which are divisible by 3 or 5 are 10 but if you look in both then you will see 15 is common in both cases so total numbers which are divisible 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18, 20 mean 9.
Hence the answer is 9/20.
From 1 to 20 there divisible by 3 is -> 3, 6, 9, 12, 15, 18 and divisible by 5 is -> 5, 10, 15, 20.
Now if you see total numbers which are divisible by 3 or 5 are 10 but if you look in both then you will see 15 is common in both cases so total numbers which are divisible 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18, 20 mean 9.
Hence the answer is 9/20.
(8)
Ganapathithanuja said:
10 months ago
If 15 is there then the answer is 1/2.
Am I right? Anyone, please explain.
Am I right? Anyone, please explain.
(29)
Jasleen said:
10 months ago
I can't understand this question⁉️
Please help me.
Please help me.
(5)
Ephrem Alemu Mantose ETH said:
1 year ago
The answer is correct.
P(A U B) = P(A) + P(B) - P(AnB),
= 6/20 + 4/20 - 1/20,
= 9/20.
P(A U B) = P(A) + P(B) - P(AnB),
= 6/20 + 4/20 - 1/20,
= 9/20.
(15)
Tanya said:
1 year ago
Yes, 9/20 is the correct option because 15 is a multiple of both 3 and 5.
So we considered it only once (either with 3 or either with 5).
So we considered it only once (either with 3 or either with 5).
(23)
PURVA said:
2 years 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}.
(19)
Sriram said:
2 years 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.
(35)
Abhishek tomar said:
2 years ago
15 is the multiple of 3 and 5. while question is about the multiple of 3 or 5 so, why not option C?
(7)
Niraml said:
2 years 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.
(2)
Avaneet Agrahari said:
4 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
(11)
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