Question

A box contains $90$ discs which are numbered from $1$ to $90$. If one disc is drawn at random from the box, find the probability that it bears (i) a two digit number (ii) a perfect square number (iii) a number divisible by $5$.

Answer 1

Solution(i):

Let $E$ be the event of drawing a disc from the box of discs numbered from $1$ to $90$

Two digit numbers from $1$ to $90=10,11,12,.....,90$
No. of favorable outcomes$=81$

Total no. of possible outcomes $=90$

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{81}{90}$$=\frac{9}{10}$

Therefore, the probability of a two digit numbered disc from the box of discs numbered from $1$ to $90$$=\frac{9}{10}$

Solution(ii):

Let $F$ be the event of drawing a perfect square numbered disc from the box of discs numbered from $1$ to $90$

Perfect square numbers from $1$ to $90=1,4,9,16,25,36,49,64,81$
No. of favorable outcomes$=9$

Total no. of possible outcomes $=90$

We know that, Probability $P\left(F\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{9}{90}$$=\frac{1}{10}$

Therefore, the probability of drawing a perfect square numbered disc from the box of discs numbered from $1$ to $90$$=\frac{1}{10}$

Solution(iii):

Let $G$ be the event of drawing a disc with a number divisible by $5$ from the box of discs numbered from $1$ to $90$

Numbers divisible by $5$ from $1$ to $90=5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90$
No. of favorable outcomes$=18$

Total no. of possible outcomes $=90$

We know that, Probability $P\left(G\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{18}{90}$$=\frac{1}{5}$

Therefore, the probability of drawing a disc with a number divisible by $5$ from the box of discs numbered from $1$ to $90$$=\frac{1}{5}$