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 4 of 11.
Santosh said:
1 decade ago
Yah by using P(AUB)formula we can calculate this, So no doubt persists-P(A)=6/20, P(B)=4/20,and P(A^B)=1/20(since 15 common for 3,5),
Finally answer is 6/20+4/20-1/20 = 9/20.
Finally answer is 6/20+4/20-1/20 = 9/20.
Mosca said:
1 decade ago
Everyone is wrong. The answer is D: 9/20.
I am the best at maths see. This is because it is out of 20 and the multiplier of 3 and 5 is 9.
You see? So, it has to be 9/20.
I am the best at maths see. This is because it is out of 20 and the multiplier of 3 and 5 is 9.
You see? So, it has to be 9/20.
Lalitha said:
7 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?
Adhamki dileep kumar said:
1 decade ago
Sample space p(s) = 20c1 = 20.
Events occurred p(e) = (3,6,9,12,15,18,5,10,15,20) = 10.
= (3,6,9,12,15,18,5,10,20)=9
Probability = p(e)/p(s).
= 9/20.
Events occurred p(e) = (3,6,9,12,15,18,5,10,15,20) = 10.
= (3,6,9,12,15,18,5,10,20)=9
Probability = p(e)/p(s).
= 9/20.
Niraml said:
3 years ago
Here, the total ticket is 20 we gonna take 1 from that 20 tickets it can be multiple of 3 or 5
so,
5C1 + 3C1/20C1 = 5+3/20 = 8/20.
= 2/5 is correct.
so,
5C1 + 3C1/20C1 = 5+3/20 = 8/20.
= 2/5 is correct.
(2)
Sohail said:
1 decade ago
15 creates a big confusion.
Because my answer is 1/2...:(
p(3)={3,6,9,12,15,18}
p(5)={5,10,15,20}
P(3u5)=p(3)+p(5)
=6/20+4/20
=6+4/20
=10/20
=1/2
Because my answer is 1/2...:(
p(3)={3,6,9,12,15,18}
p(5)={5,10,15,20}
P(3u5)=p(3)+p(5)
=6/20+4/20
=6+4/20
=10/20
=1/2
PURVA said:
3 years ago
@All.
Here is my solution.
P(A or B) = P(A) + P(B) -P(A AND B).
P(3) = 6/20 {3,6,9,12,15,18},
P(5) = 4/20 {5,10,15,20},
P(3 and 5) = 1/20 {15}.
Here is my solution.
P(A or B) = P(A) + P(B) -P(A AND B).
P(3) = 6/20 {3,6,9,12,15,18},
P(5) = 4/20 {5,10,15,20},
P(3 and 5) = 1/20 {15}.
(21)
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).
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.
Garima said:
10 years ago
According to me the answer should be 1/2.
Multiples of 3 and 5. (1, 3, 5, 6, 9, 10, 12, 15, 18, 20).
Please let me know where am I wrong?
Multiples of 3 and 5. (1, 3, 5, 6, 9, 10, 12, 15, 18, 20).
Please let me know where am I wrong?
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