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:
107 comments Page 3 of 11.
Sri said:
6 years ago
@All.
Hi, Why don't we get 15?
It is a multiple of 5.
So P(E)={3,6,9,12,15,18,5,10,15,20}. So these are the multiples of 3 and 5. Total is 10. So P(E)/P(S) = 10/20 .That is 1/2.
Hi, Why don't we get 15?
It is a multiple of 5.
So P(E)={3,6,9,12,15,18,5,10,15,20}. So these are the multiples of 3 and 5. Total is 10. So P(E)/P(S) = 10/20 .That is 1/2.
Sagaana said:
6 years ago
How to solve the same problem when the 1000 tickets mixed in the pack what is the probability of getting multiples of 3 and 5?
V S Abhirame said:
6 years ago
@Souravbaidya.
Circular board of i feet has area Pi. The area of the region containing all points which are closed to center than to periphery will be Pi/4. Hence the proportion to hit near the center is 1/4=.25. So, D is the correct answer.
Circular board of i feet has area Pi. The area of the region containing all points which are closed to center than to periphery will be Pi/4. Hence the proportion to hit near the center is 1/4=.25. So, D is the correct answer.
Siva said:
6 years ago
How come 4/20 & 6/20?
Asiya said:
6 years ago
My answer is 8/20 because in question it is stated that we take multiple of 3 or 5 not multiple of both so we should neglect 15 because it is the multiple of 3 and 5.
Anyone please help me to get it.
Anyone please help me to get it.
Deepika Balaji said:
7 years ago
Can anyone help me in solving this problem. What is the probability of getting at least one six in a single throw of three unbiased dice?
Azoya said:
7 years ago
Multiple of 3 = {3,6,9,12,15,18},
Multiple of 5 ={5,10,15},
Multiple of both 3 and 5 = {15},
So p(3) = 6/20,
p(5) = 3/20,
p(3 and 5) = 1/20.
So, answer is = 6/20 + 3/20 -1/20.
i,e. 8/20 which is 2/5 according to options.
I am getting this answer because the question is multiple of 3 or 5 . So we need to exclude multiple of 3 and 5.
Multiple of 5 ={5,10,15},
Multiple of both 3 and 5 = {15},
So p(3) = 6/20,
p(5) = 3/20,
p(3 and 5) = 1/20.
So, answer is = 6/20 + 3/20 -1/20.
i,e. 8/20 which is 2/5 according to options.
I am getting this answer because the question is multiple of 3 or 5 . So we need to exclude multiple of 3 and 5.
Anonymous said:
7 years ago
Why didnt we apply the (not A) formula?
i.e. (Not blue ball) = 1 - (blue ball).
= 1 - 2/7.
= 5/7.
i.e. (Not blue ball) = 1 - (blue ball).
= 1 - 2/7.
= 5/7.
Goutam said:
7 years ago
9/20 is correct answer.
5 multiplies are 5,10,15,20.
3 multiplies are 3,9,12,15,18.
Answer = 9/20.
5 multiplies are 5,10,15,20.
3 multiplies are 3,9,12,15,18.
Answer = 9/20.
Thrideep said:
7 years ago
I think The answer is 3/4.
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