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# 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?

A. $$\frac{{1}}{{2}}$$

B. $$\frac{{2}}{{5}}$$

C. $$\frac{{8}}{{15}}$$

D. $$\frac{{9}}{{20}}$$

### Solution(By Examveda Team)

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}
$$\therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{9}{{20}}$$

This Question Belongs to Arithmetic Ability >> Probability

1. Okay I got it

2. What is multiple of a number?

3. Hw 9 comes

4. How

5. This is a Joint Probability question and formula to calculate joint probability is :
P(A or B) = P(A) + P(B) - P(A and B)
Here P(A) = Value of Multiple of 3 are {3,6,9,12,15,18} / Sample space = 6/20
P(B) = Value of Multiple of 5 are {5,10,15,20} / Sample space = 4/20
P(A and B) = Common favorable events between A and B are {15} = 1/20

As per formula P(A or B) = 6/20 + 4/20 - 1/20 = 9/20

6. Where is 15? multiple of 5

7. 5 will have to be there, and then the answer would be 2/5.
Why don't you guys just correct it.

8. 3 is not a multiple of 3. so why 3 is here. And if it is why 5 is not?

9. KINDLY EXPLAIN MORE

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