Answer: Option B

Answer : Option B

Explanation :

---------------------------------------------------------------------------------------
Solution 1
---------------------------------------------------------------------------------------
Total number of tickets, n(S)= 20
To get a multiple of 3 , the favorable cases are 3,9,6,12,15,18.
=> Number of ways in which we get a multiple of 3 = 6

$MF#%\text{P(Multiple of 3) = }\dfrac{\text{Number of ways in which we get a sum of multiple of 3}}{\text{Total number of outcomes possible}} = \dfrac{6}{20}$MF#%

To get a multiple of 5, the favorable cases are 5,10,15,20.
=> Number of ways in which we get a multiple of 5 = 4

$MF#%\text{P(Multiple of 5) = }\dfrac{\text{Number of ways in which we get a multiple of 5}}{\text{Total number of outcomes possible}} = \dfrac{4}{20}$MF#%

There are some cases where we get multiple of 3 and 5. the favorable case for this is 15
=> Number of ways in which we get a multiple of 3 and 5 = 1

$MF#%\text{P(Multiple of 3 and 5) = }\dfrac{\text{Number of ways in which we get a multiple of 3 and 5}}{\text{Total number of outcomes possible}} = \dfrac{1}{20}$MF#%

Here a number can be both a a multiple of 3 and 5. Hence these are not mutually exclusive events.
, we have
P(multiple of 3 or 5) = P(multiple of 3) + P(multiple of 5)- P(multiple of 3 and 5)

$MF#%= \dfrac{6}{20} + \dfrac{4}{20} - \dfrac{1}{20} = \dfrac{9}{20}$MF#%

--------------------------------------------------------------------------------------- Solution 2 --------------------------------------------------------------------------------------- Total number of tickets, n(S)= 20
To get a multiple of 3 or 5, the favorable cases are 3, 5, 6, 9, 10, 12,15,18, 20.
=>Number of ways in which we get a multiple of 3 or 5 = 9

$MF#%\text{P(multiple of 3 or 5) = }\dfrac{\text{Number of ways in which we get a multiple of 3 or 5}}{\text{Total number of outcomes possible}} = \dfrac{9}{20}$MF#%