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?
1
2
2
5
8
15
9
20
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 2 of 11.

Rajesh Bhadra said:   2 months ago
@Ganapathithanuja.

From 1 to 20 there divisible by 3 is -> 3, 6, 9, 12, 15, 18 and divisible by 5 is -> 5, 10, 15, 20.

Now if you see total numbers which are divisible by 3 or 5 are 10 but if you look in both then you will see 15 is common in both cases so total numbers which are divisible 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18, 20 mean 9.

Hence the answer is 9/20.
(3)

Anitha said:   6 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.
(2)

Niraml said:   2 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.
(2)

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.
(1)

Veeru said:   1 decade ago
Sundar & Manju thought wrong because 15 already be consider.

So ans will be 9/20.
(1)

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.
(1)

Dabir Masood said:   8 years ago
Yes I too agree that 1/2 is the correct answer for this question.
(1)

Uday said:   5 years ago
(3, 6,9,12,15,18,5,10,15,20).
= 10/20,
= 1/2.
(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.

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).


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