Question

A bag contains tickets numbered $11,12,13,.....30$. A ticket is taken out from the bag at random. Find the probability that the number on the drawn ticket (i) is a multiple of $7$ (ii) is greater than $15$ and a multiple of $5$.

Answer 1

Solution(i):

Let $E$ be event of getting a multiple of $7$ from the numbers $11$ to $30$

Multiples of $7$ from $11$ to $30=14,21,28$
No. of favorable outcomes$=3$

Total no. of possible outcomes $=20$

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{3}{20}$

Therefore, the probability that a number selected from the numbers $11,12,13,....,30$ is a multiple of $7$ $=\frac{3}{20}$

Solution(i):

Let $E$ be event of getting a no. greater than $15$ and a multiple of $5$ from the numbers $11$ to $30$

Nos. greater than $15$ and multiple of $5$ from the numbers $11$ to $30(=20,25,30)$
No. of favorable outcomes$=3$

Total no. of possible outcomes $=20$

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{3}{20}$

Therefore, the probability that a number selected from the numbers $11,12,13,....,30$ is greater than $15$ and multiple of $5$ is $\frac{3}{20}$