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:
106 comments Page 1 of 11.
Sundar said:
1 decade ago
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, 15, 20}.
P(E) = 1/2.
Is this correct? Please give your thoughts on this.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
P(E) = 1/2.
Is this correct? Please give your thoughts on this.
(1)
Karthik said:
1 decade 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.
Veeru said:
1 decade ago
Sundar & Manju thought wrong because 15 already be consider.
So ans will be 9/20.
So ans will be 9/20.
(1)
Sneha 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).
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.
Souravbaidya said:
1 decade ago
A circular dartboard of radius 1 foot is at a distance of 20 feet from you. You throw a dart at it and it
hits the dartboard at some point Q in the circle. What is the probability that Q is closer to the center
of the circle than the periphery?
a) 0.75 b) 1 c) 0.5 d) 0.25
hits the dartboard at some point Q in the circle. What is the probability that Q is closer to the center
of the circle than the periphery?
a) 0.75 b) 1 c) 0.5 d) 0.25
Pawan said:
1 decade ago
Hi Guys ..
according to me the answer is 2/5
A = multiple of 3
B = multiple of 5
P(A U B) = P(A) + P(B) - P(A n B)
n(A) = 6 ( 3,6,9,12,15,18)
n(B) = 3 (5,10,15)
n (A n B) = 1 (15 --> this comes in both)
Hence, prob = 6/20 + 3/20 - 1/20 = 8/20 = 2/5
according to me the answer is 2/5
A = multiple of 3
B = multiple of 5
P(A U B) = P(A) + P(B) - P(A n B)
n(A) = 6 ( 3,6,9,12,15,18)
n(B) = 3 (5,10,15)
n (A n B) = 1 (15 --> this comes in both)
Hence, prob = 6/20 + 3/20 - 1/20 = 8/20 = 2/5
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.
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)
Satyendra Shukla said:
1 decade ago
Answer will be 9/20, because 15 already be considered, so it can not be consider next time.
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