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 8 of 11.
Rutu said:
9 years ago
Why s=20 selected? please give me answer.
Mahesh said:
9 years ago
What's, and what is e please explain?
Shailendra rathore said:
9 years ago
9/20 is the right answer, and thanks for explaining the answer in different ways.
Saurav said:
9 years ago
As both the events A and B are independent and non-mutually exclusive we can use the following formula:.
P (A or B) = P (A) + P (B) - P (A and B).
= 6/20 + 4/20 - (6/20 * 4/20) [probability intersection rule for dependent events].
= 10/20 - 3/50.
= 50 - 6/100.
= 44/100 = 11/25. But this answer is not in the option.
P (A or B) = P (A) + P (B) - P (A and B).
= 6/20 + 4/20 - (6/20 * 4/20) [probability intersection rule for dependent events].
= 10/20 - 3/50.
= 50 - 6/100.
= 44/100 = 11/25. But this answer is not in the option.
Rani said:
9 years ago
Where this 5 gone in the event?
Mani said:
9 years ago
Where this 6 gone in the event?
Nicole said:
9 years ago
I'm confused, on how thy got D as an answer?
Gad said:
9 years ago
A = multiple of 3 (3, 6, 9, 12, 15, 18).
Possible outcome = 6.
B = multiple of 5 (5, 10, 15, 20).
Possible outcome = 4.
Therefore, P (A or B) = P (A) + P (B).
6/20 + 4/20 = 10/20 = 1/2.
Possible outcome = 6.
B = multiple of 5 (5, 10, 15, 20).
Possible outcome = 4.
Therefore, P (A or B) = P (A) + P (B).
6/20 + 4/20 = 10/20 = 1/2.
Anonymous said:
9 years ago
The correct answer is 9/20.
Deva Bala said:
9 years ago
The option d) 9/20 is the correct answer.
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