# 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:

103 comments Page 2 of 11.
Shivu said:
5 years ago

Ans is 9/20 because;

Multiple of 3 (3 6 9 12 15 18)=possiblity 6 so 6/20,

Multiple of 5 (5 10 15 20)=possibility 4 so 4/20,

15 is both sides so eliminate, so -1/20.

6/20+4/20-1/20 = 9/20.

Multiple of 3 (3 6 9 12 15 18)=possiblity 6 so 6/20,

Multiple of 5 (5 10 15 20)=possibility 4 so 4/20,

15 is both sides so eliminate, so -1/20.

6/20+4/20-1/20 = 9/20.

(4)

Anitha said:
5 years ago

@Shraddha.

5 is also considered so that we get probability 9/20. If 5 is not considered we can't get 9/20.

5 is also considered so that we get probability 9/20. If 5 is not considered we can't get 9/20.

(2)

Shraddha said:
5 years ago

But why is the number 5 not considered? 5 is a multiple of 5!

Lalitha said:
5 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?

Ramesh said:
5 years ago

Multiples of 3 = 3,6,9,12,15,18 ----> event A.

Multiples of 5 = 5,10,15,20 ------> event B.

Both are having 15 as common so these are dependent events.

P(A)= 6/20.

P(B)= 4/20.

P(AUB)= P(A)+P(B)-P(A intersection B).

= P(A)+P(B)-P(A)*P(B/A),

= 6/20 + 4/20 - {(6/20)*(1/6)}.

= 9/20.

Multiples of 5 = 5,10,15,20 ------> event B.

Both are having 15 as common so these are dependent events.

P(A)= 6/20.

P(B)= 4/20.

P(AUB)= P(A)+P(B)-P(A intersection B).

= P(A)+P(B)-P(A)*P(B/A),

= 6/20 + 4/20 - {(6/20)*(1/6)}.

= 9/20.

Tushar said:
5 years ago

Among the 20 numbers multiples of 3=20/3 which is 6. And similarly, multiples of 5 are=20/5 which are 4. Now total multiples are 10. Now among these 10 multiples, we have one common multiple that is 15.

Now remaining multiples=total multiples -common multiple.

=10-1.

=9.

Now required prob =9/20.

Now remaining multiples=total multiples -common multiple.

=10-1.

=9.

Now required prob =9/20.

Sri said:
5 years ago

@All.

Hi, Why don't we get 15?

It is a multiple of 5.

So P(E)={3,6,9,12,15,18,5,10,15,20}. So these are the multiples of 3 and 5. Total is 10. So P(E)/P(S) = 10/20 .That is 1/2.

Hi, Why don't we get 15?

It is a multiple of 5.

So P(E)={3,6,9,12,15,18,5,10,15,20}. So these are the multiples of 3 and 5. Total is 10. So P(E)/P(S) = 10/20 .That is 1/2.

Sagaana said:
5 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?

V S Abhirame said:
5 years ago

@Souravbaidya.

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.

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.

Siva said:
5 years ago

How come 4/20 & 6/20?

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