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 \[\begin{align} 5 & =\frac{\text{ Number of possible outcomes }}{\text{No of favourable outcomes}} \\ {} & =\frac{18}{90} \\ {} & =\frac{1}{5} \\ \end{align}\]