Question

A box contains $90$ discs which are numbered from $1to90$. If one disc is drawn at random from the box, find the probability that it bears a two-digit number

Answer 1

Step $1$: Find the total number of outcomes and favorable number of outcomes

The total number of discs$=90$

Since, $1$ to $9$ are single digit numbers.

So, total $2$ digit numbers are $90-9=81$

Total number of discs having two-digit numbers$=81$

Step 2: Find the required probability using the classical definition of probability

$\mathrm{P}\left(\mathrm{E}\right)=\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}$

$\begin{array}{rcl}\mathrm{P}(\mathrm{bearing}\mathrm{a}\mathrm{two}\mathrm{digit}\mathrm{number})& =& \frac{81}{90}\\ & =& \frac{9}{10}\\ & =& 0.9\end{array}$

Hence, the probability of getting a two-digit number is $0.9$