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

We have to find the probability for the required options. Probability is a type of ratio where we compare how many times an outcome can occur compared to all possible outcomes.
${\text{Probability = }}\dfrac{{\left( {{\text{The number of wanted outcomes}}} \right)}}{{\left( {{\text{The number of possible outcomes}}} \right)}}$

Complete step-by-step answer:
It is given that a box contains $90$ discs which are numbered from $1$ to $90$.
One disc is drawn at random from the box.
(i). We need to determine the probability that it bears a two-digit number.
Two digit numbers are from $10$ to $90$ which are $81$ in number.
Total number of discs =$90$
So the possible outcome of choosing one disc at random from the box, is $^{90}{C_1}$ .
Thus the possible outcome of choosing one disc and it bears a two-digit number is $^{81}{C_1}$.
The probability that one disc is taken out is a two digit number
$\Rightarrow \dfrac{{\left( {{\text{The number of wanted outcomes}}} \right)}}{{\left( {{\text{The number of possible outcomes}}} \right)}}$
$\Rightarrow \dfrac{{^{81}{C_1}}}{{^{90}{C_1}}}$
Using the combination formula,
$\Rightarrow \dfrac{{\dfrac{{81!}}{{1!80!}}}}{{\dfrac{{90!}}{{1!89!}}}}$
Simplifying the above terms we get,
$\Rightarrow \dfrac{{\dfrac{{81 \times 80!}}{{80!}}}}{{\dfrac{{90 \times 89!}}{{89!}}}}$
$\Rightarrow \dfrac{{81}}{{90}}$
Thus we finally get the probability that one disc is taken out as a two digit number is $\dfrac{{81}}{{90}}$.
(ii). We need to determine the probability that it bears a perfect square number.
The perfect square numbers between $1$ to $90$ are $1,4,9,16,25,36,49,64,81$ which are $9$ in number.
Total number of discs $= 90$
So the possible outcome of choosing one disc at random from the box, is $^{90}{C_1}$ .
Thus the possible outcome of choosing one disc and it bears a perfect square number, is $^9{C_1}$.
The probability that one disc is taken out is a perfect square number
$\Rightarrow \dfrac{{\left( {{\text{The number of wanted outcomes}}} \right)}}{{\left( {{\text{The number of possible outcomes}}} \right)}}$
$\Rightarrow \dfrac{{^9{C_1}}}{{^{90}{C_1}}}$
Using the combination formula,
$\Rightarrow \dfrac{{\dfrac{{9!}}{{1!8!}}}}{{\dfrac{{90!}}{{1!89!}}}}$
Simplifying the above term we get,
$\Rightarrow \dfrac{{\dfrac{{9 \times 8!}}{{8!}}}}{{\dfrac{{90 \times 89!}}{{89!}}}}$
\Rightarrow \dfrac{9}{{90}} = \dfrac{1}{{10}}$$$$$$ The probability that one disc is taken out as a perfect square number is\dfrac{1}{{10}}$. (iii). We need to determine the probability that it bears a number divisible by$5$. The numbers divisible by$5$between$1$to$90$are$5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90$which are$18$in number. Total number of discs =$90$So the possible outcome of choosing one disc at random from the box, is$^{90}{C_1}$. Thus the possible outcome of choosing one disc and it bears a number divisible by$5$, is$^{18}{C_1}$. The probability that one disc is taken out is a number divisible by$5$$ \Rightarrow \dfrac{{\left( {{\text{The number of wanted outcomes}}} \right)}}{{\left( {{\text{The number of possible outcomes}}} \right)}}$$ \Rightarrow \dfrac{{^{18}{C_1}}}{{^{90}{C_1}}}$Using the combination formula,$ \Rightarrow \dfrac{{\dfrac{{18!}}{{1!17!}}}}{{\dfrac{{90!}}{{1!89!}}}}$Simplifying the above term we get,$ \Rightarrow \dfrac{{\dfrac{{18 \times 17!}}{{17!}}}}{{\dfrac{{90 \times 89!}}{{89!}}}}$$ \Rightarrow \dfrac{{18}}{{90}} = \dfrac{1}{5}$$$$$$
The probability that one disc is taken out is a number divisible by $5$ is $\dfrac{1}{5}$.

Note: A combination is a grouping or subset of items.
For a combination,
$C{\left( {n,r} \right)^n} = {C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Where, factorial n is denoted by $n!$ and defined by
$n! = n(n - 1)(n - 2)(n - 4).......2.1$