# 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.
Rajesh Bhadra said:
1 month 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.

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)

Ganapathithanuja said:
7 months ago

If 15 is there then the answer is 1/2.

Am I right? Anyone, please explain.

Am I right? Anyone, please explain.

(22)

Jasleen said:
7 months ago

I can't understand this question⁉️

Please help me.

Please help me.

(5)

Ephrem Alemu Mantose ETH said:
11 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.

(10)

Tanya said:
12 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).

(21)

PURVA said:
1 year 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}.

(19)

Sriram said:
2 years 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.

(34)

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?

(7)

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.

so,

5C1 + 3C1/20C1 = 5+3/20 = 8/20.

= 2/5 is correct.

(2)

Avaneet Agrahari said:
4 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

(11)

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