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 6 of 11.
Ketan said:
1 decade ago
p(3) = {3,6,9,12,15,18}.
p(5) = {5,10,15,20}.
P(3u5) = p(3)+p(5).
I don't understand how it's come
= 6/20+4/20 ?
p(5) = {5,10,15,20}.
P(3u5) = p(3)+p(5).
I don't understand how it's come
= 6/20+4/20 ?
Keerthi said:
1 decade ago
Here he said 3 or 5 but not 3&5 so answer will be 1\2.
Since or means we should add &and means multiply.
Since or means we should add &and means multiply.
Shekar said:
1 decade ago
A(n)={3,6,9,12,15,18}
B(n)={5,10,15,20}
P(AuB)=P(A)+P(B)-P(A intersection B)
P(AuB)=6/20+4/20-1/20
P(AuB)=9/20.
B(n)={5,10,15,20}
P(AuB)=P(A)+P(B)-P(A intersection B)
P(AuB)=6/20+4/20-1/20
P(AuB)=9/20.
Parthasarathy said:
1 decade ago
A(n)={3,6,9,12,15,18}
B(n)={5,10,15,20}
P(AuB)=P(A)+P(B)-P(A intersection B)
P(AuB)=6/20+4/20-1/20
P(AuB)=9/20.
B(n)={5,10,15,20}
P(AuB)=P(A)+P(B)-P(A intersection B)
P(AuB)=6/20+4/20-1/20
P(AuB)=9/20.
Sifuna charles said:
1 decade ago
Shekar your explanation is good but you should have included the intersection part of it for easy understanding.
Avaneet Agrahari said:
5 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
(12)
Divya said:
10 years ago
How to cal multiple of larger number?
Can any one give a answer for this?
Ex: Same question number is 1-300.
Can any one give a answer for this?
Ex: Same question number is 1-300.
Sugnya said:
1 decade ago
The answer is 1/2 because if there is same number we should include it. This not union or intersection model.
Anitha said:
7 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.
(2)
Ujjwal shukla said:
9 years ago
Find the probability that a number from 1 to 300 is divisible by 3 or 7.
Can anyone answer this question.
Can anyone answer this question.
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