# Ex.15.1 Q18 Probability Solution - NCERT Maths Class 10

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## 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 $$5$$.

## Text Solution

What is known?

Total number of discs = $$90$$

Total number of 2digit numbers between $$1$$ to $$90$$ = $$81$$

Total number of perfect square numbers between $$1$$ to $$90$$ are $$1,4,9,16,25,36,49,64,81= 9$$

Total numbers that are divisible by $$5$$ are $$5$$,$$10$$,$$15$$,$$20$$,$$25$$,$$30$$,$$35$$,$$40$$,$$45$$,$$50$$,$$55$$,$$60$$,$$65$$,$$70$$,$$75$$,$$80$$,$$85$$,$$90$$$$=18$$

What is unknown?

The probability that the disc bears

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by $$5$$.

Steps:

(i) Probability of getting a two digit number \begin{align} & =\frac{\text{ Number of possible outcomes}}{\text{Total no of favourable outcomes}} \\ {} & =\frac{81}{90}\\&=\frac{9}{10} \\\end{align}

(ii) Probability of getting a perfect square number \begin{align} & =\frac{\text{ Number of possible outcomes}}{\text{Total no of favourable outcomes}} \\& =\frac{9}{90}\\&=\frac{1}{10} \\ \end{align}

(iii) Probability of getting a number divisible by \begin{align} 5 & =\frac{\text{ Number of possible outcomes }}{\text{No of favourable outcomes}} \\ {} & =\frac{18}{90} \\ {} & =\frac{1}{5} \\ \end{align}

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