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

## 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

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorabl outcomes} \end{bmatrix} } \\& =\frac{81}{90}\\&=\frac{9}{10} \end{align}\]

(ii) Probability of getting a perfect square number

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\ & =\frac{9}{90}\\&=\frac{1}{10} \end{align}\]

(iii) Probability of getting a number divisible by \(5\)

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\ & =\frac{18}{90} \\ & =\frac{1}{5} \end{align}\]