There are 25 tickets numbered 1, 2, 3, 4,…, 25 respectively. One ticket is draw at random.
Question:

There are 25 tickets numbered 1, 2, 3, 4,…, 25 respectively. One ticket is draw at random. What is the probability that the number on the ticket is a multiple of 3 or 5?

(a) $\frac{2}{5}$

(b) $\frac{11}{25}$

(C) $\frac{12}{25}$

 

(d) $\frac{13}{25}$

 

Solution:

(c) $\frac{12}{25}$

Explanation:
​Total number of tickets = 25
Let E be the event of getting a multiple of 3 or 5.
Then,
Multiples of 3 are 3, 6, 9, 12, 15, 18, 21 and 24.
Multiples of 5 are 5, 10, 15, 20 and 25.

Number of favourable outcomes = ( 8 + 5 − 1) = 12      

$\therefore P($ getting a multiple of 3 or 5$)=P(\mathrm{E})=\frac{12}{25}$

 

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