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 5 of 11.
Dipesh said:
1 decade ago
Yes, answer is 9/20. We should not consider a number more than one time (15 is a multiple of 3 and 5. So, it cannot be used more than once)
Isha said:
8 years ago
Please help me to solve this-
A number is selected at random from first 50 number. Find the probability that it is a multiple of 3 and 4.
A number is selected at random from first 50 number. Find the probability that it is a multiple of 3 and 4.
Arun said:
1 decade ago
I can't understand.
P (e) ={3, 6, 9, 12, 15, 18, 5, 10, 20}.
We multiplying p (e) with s then we don't want to multiply s={7, 8 to 18}.
P (e) ={3, 6, 9, 12, 15, 18, 5, 10, 20}.
We multiplying p (e) with s then we don't want to multiply s={7, 8 to 18}.
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?
Tanya said:
2 years 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).
(26)
Karthik said:
2 decades ago
No. You are wrong Sundar.
Because, 15 has already consider once so there is no need to consider it again.
So the answer is 9/20.
Because, 15 has already consider once so there is no need to consider it again.
So the answer is 9/20.
Naina said:
8 years ago
A coin is tossed live times. What is the probability that there is at the least one tail?
Can anyone help me with this question?
Can anyone help me with this question?
Praveen Kumar Gautam said:
1 decade ago
Answer will be 9/20, because 15 already be considered, so it can not be consider next time. So Sundar & Manju thought wrong.
Sagaana said:
7 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?
ANIL CHAUHAN 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.
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