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 2 of 11.
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)
Kavya said:
5 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.
(5)
Kousalya said:
5 years ago
Yes, I too agree 1/2 is the right answer.
Uday said:
5 years ago
(3, 6,9,12,15,18,5,10,15,20).
= 10/20,
= 1/2.
= 10/20,
= 1/2.
(1)
Shivu said:
6 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.
(6)
Anitha said:
6 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)
Shraddha said:
6 years ago
But why is the number 5 not considered? 5 is a multiple of 5!
Lalitha said:
6 years ago
If the ticket numbers are taken from 1 2 3 ..... 100 what's the probability of getting the probability of card which multiple by 3 or 5. Can anyone solve this?
Ramesh said:
6 years ago
Multiples of 3 = 3,6,9,12,15,18 ----> event A.
Multiples of 5 = 5,10,15,20 ------> event B.
Both are having 15 as common so these are dependent events.
P(A)= 6/20.
P(B)= 4/20.
P(AUB)= P(A)+P(B)-P(A intersection B).
= P(A)+P(B)-P(A)*P(B/A),
= 6/20 + 4/20 - {(6/20)*(1/6)}.
= 9/20.
Multiples of 5 = 5,10,15,20 ------> event B.
Both are having 15 as common so these are dependent events.
P(A)= 6/20.
P(B)= 4/20.
P(AUB)= P(A)+P(B)-P(A intersection B).
= P(A)+P(B)-P(A)*P(B/A),
= 6/20 + 4/20 - {(6/20)*(1/6)}.
= 9/20.
Tushar said:
6 years ago
Among the 20 numbers multiples of 3=20/3 which is 6. And similarly, multiples of 5 are=20/5 which are 4. Now total multiples are 10. Now among these 10 multiples, we have one common multiple that is 15.
Now remaining multiples=total multiples -common multiple.
=10-1.
=9.
Now required prob =9/20.
Now remaining multiples=total multiples -common multiple.
=10-1.
=9.
Now required prob =9/20.
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