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 10 of 11.
Sreejith said:
1 decade ago
Hello Rach,
How did you get that as pi/4?
Or anybody else knows?
How did you get that as pi/4?
Or anybody else knows?
Rahul said:
1 decade ago
@rach.
Hai rach I didn't get you. Can you please explain clearly.
Hai rach I didn't get you. Can you please explain clearly.
Rach said:
1 decade ago
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.
Kiruba said:
1 decade ago
Thanks sneha. Ans is 9/20.
Albert said:
1 decade ago
Final answer is 9/20
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.
Dharmesh Patel said:
1 decade ago
Yes answer is Only and only 9/20.
Satyendra Shukla said:
1 decade ago
Answer will be 9/20, because 15 already be considered, so it can not be consider next time.
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)
Murugesh said:
1 decade ago
Hi @pawan,
according to your assumption, you had missed the 20 in n(B)
so n(B)={5,10,15,20}
then P(B)=4/20
Hence P(AUB)=6/20+4/20-1/20=9/20.
according to your assumption, you had missed the 20 in n(B)
so n(B)={5,10,15,20}
then P(B)=4/20
Hence P(AUB)=6/20+4/20-1/20=9/20.
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