Question

A bag contains cards numbered from $1$ to $49$. A card is drawn from the bag at random, after mixing the card thoroughly. Find the probability that the number on the drawn card is
(i) an odd number (ii) a multiple of $5$ (iii) a perfect square (iv) an even prime number

Answer 1

Solution(i):

Let $E$ be the event of drawing an odd-numbered card from the cards numbered $1$ to $49$

Odd numbers from $1$ to $49=1,3,5,.....,49$
No. of favorable outcomes$=25$

Total no. of possible outcomes $=49$

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{25}{49}$

Therefore, the probability of an odd number card from the cards numbered $1$ to $49$$=\frac{25}{49}$

Solution(ii):

Let $G$ be the event of drawing a number - multiple of $5$ from the cards numbered from $1$ to $49$

Multiples of $5$ from $1$ to $49=5,10,15,20,25,30,35,40,45$
No. of favorable outcomes$=9$

Total no. of possible outcomes $=49$

We know that, Probability $P\left(G\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{9}{49}$

Therefore, the probability of drawing a multiple of $5$ from the cards numbered from $1$ to $49$$=\frac{9}{49}$

Solution(iii):

Let $F$ be the event of drawing a perfect square number from the cards numbered from $1$ to $49$

Perfect square numbers from $1$ to $49=1,4,9,16,25,36,49$
No. of favorable outcomes$=7$

Total no. of possible outcomes $=49$

We know that, Probability $P\left(F\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{7}{49}$$=\frac{1}{7}$

Therefore, the probability of drawing a perfect square number from the cards numbered from $1$ to $49$$=\frac{1}{7}$

Solution(iv):

Let $H$ be the event of drawing an even prime number from the cards numbered $1$ to $49$

Even prime numberfrom $1$ to $49=2$ $(2$ is the only even prime number $)$
No. of favorable outcomes$=1$

Total no. of possible outcomes $=49$

We know that, Probability $P\left(H\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{1}{49}$

Therefore, the probability of drawing an even prime number from the cards numbered $1$ to $49$$=\frac{1}{49}$