# 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 1 of 11.
Sriram said:
12 months ago

The multiple of 3 and 5 is given numbers 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,7,18,19,20. While the question is about the multiple of 3 is 3,6,9,12,15,18 and the multiple of 5 is 5,10,15,20.

The total number is 10.

So,

10/20 = 1/2.

The total number is 10.

So,

10/20 = 1/2.

(29)

PURVA said:
10 months 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}.

(14)

Avaneet Agrahari said:
3 years ago

Multiple of 3=[,3,6,912,15,18]=6

M....o. 5=[5,10,15,20]=4 but 15 is already in 3 so it count 3

Ans= 6+3/20=9/20

M....o. 5=[5,10,15,20]=4 but 15 is already in 3 so it count 3

Ans= 6+3/20=9/20

(8)

Tanya said:
4 months ago

Yes, 9/20 is the correct option because 15 is a multiple of both 3 and 5.

So we considered it only once (either with 3 or either with 5).

So we considered it only once (either with 3 or either with 5).

(7)

Abhishek tomar said:
2 years ago

15 is the multiple of 3 and 5. while question is about the multiple of 3 or 5 so, why not option C?

(6)

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)

Kavya said:
4 years ago

@Uday,

15 is repeated.

And 5 is missing.

Anyway, the answer is correct.

15 is repeated.

And 5 is missing.

Anyway, the answer is correct.

(4)

Ephrem Alemu Mantose ETH said:
4 months ago

The answer is correct.

P(A U B) = P(A) + P(B) - P(AnB),

= 6/20 + 4/20 - 1/20,

= 9/20.

P(A U B) = P(A) + P(B) - P(AnB),

= 6/20 + 4/20 - 1/20,

= 9/20.

(3)

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)

Chandan said:
1 decade ago

As per Addition Rule of Probability, we can have two set of events - Mutually Exclusive & Non-Mutually Exclusive..where Mutually Exclusive events don't occur simultaneously or together and reverse logic with Non-Mutually Exclusive events.

Having this theorem told in above para, now i come to the context:

We have 2 events,

Event A: Ticket picked is multiple of 3.

Event B: Ticket picked is multiple of 5.

Now P(A) = {3,6,9,12,15,18} = 6/20.

And P(B) = {5,10,15,20} = 4/20.

Since in P(A) & P(B) we could see 15 to be common, we could conclude that the events are Mutually Exclusive..so for such Mutually Exclusive events the Addition Rule of Probability goes like this:

P(AUB) = P(A) + P(B) - P(A n B)

Where, P(A n B) always looks for what is common in two events, which is 15 in this case, so P(A n B) = 1/20.

Therefore, P(ticket picked is multiple of 3 or 5) = 6/20 + 4/20 - 1/20 = 9/20 is the answer.

Having this theorem told in above para, now i come to the context:

We have 2 events,

Event A: Ticket picked is multiple of 3.

Event B: Ticket picked is multiple of 5.

Now P(A) = {3,6,9,12,15,18} = 6/20.

And P(B) = {5,10,15,20} = 4/20.

Since in P(A) & P(B) we could see 15 to be common, we could conclude that the events are Mutually Exclusive..so for such Mutually Exclusive events the Addition Rule of Probability goes like this:

P(AUB) = P(A) + P(B) - P(A n B)

Where, P(A n B) always looks for what is common in two events, which is 15 in this case, so P(A n B) = 1/20.

Therefore, P(ticket picked is multiple of 3 or 5) = 6/20 + 4/20 - 1/20 = 9/20 is the answer.

(1)

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