Question

Find the probability that a number selected at random from the numbers $1,2,3,......,35$ is a
(i) prime number (ii) multiple of $7$ (ii) a multiple of $3$ or $5$

Answer 1

Solution(i):

Let $E$ be event of getting a prime number from the list of numbers $1$ to $35$

Prime nos. upto $35=2,3,5,7,11,13,17,19,23,29,31$
No. of favorable outcomes$=11$

Total no. of possible outcomes $=35$

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{11}{35}$

Therefore, the probability that a prime number is selected from the numbers $1,2,3,.......,35$ is $\frac{11}{35}$

Solution(ii):

Let $F$ be event of getting a multiple of $7$ from the numbers $1$ to $35$

Multiples of $7$ upto $35=7,14,21,28,35$
No. of favorable outcomes$=5$

Total no. of possible outcomes $=35$

We know that, Probability $P\left(F\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{5}{35}=\frac{1}{7}$

Therefore, the probability that a number selected from the numbers $1,2,3,.......,35$ is a multiple of $7$ is $\frac{1}{7}$

Solution(iii):

Let $G$ be event of getting a multiple of $3$ or $5$ from the numbers $1$ to $35$

Multiples of $3$ or $5$ upto $35=3,5,6,9,10,12,15,18,20,21,24,25,27,30,33,35$

No. of favorable outcomes$=16$

Total no. of possible outcomes $=35$

We know that, Probability $P\left(G\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{16}{35}$

Therefore, the probability that a number selected from the numbers $1,2,3,.......,15$ is $\frac{16}{35}$