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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.
Given:
A box contains 90 discs which are numbered from 1 to 90.
One disc is drawn at random from the box.
To do:
We have to find the probability that it bears
(i) a two digit number.
(ii) a perfect square number.
(iii) a number divisible by 5.
Solution:
A box contains discs numbered \( 1,2,3,4, .., 89,90 \).
This implies,
The total number of possible outcomes $n=90$.
(i) Two digit numbers from 1 to 90 are $10, 11, .........., 89, 90$.
Total number of favourable outcomes $=81$.
We know that,
Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$
Therefore,
Probability that the disc bears a two digit number $=\frac{81}{90}$
$=\frac{9}{10}$
The probability that it bears a two digit number is $\frac{9}{10}$.
(ii) Perfect square numbers from 1 to 90 are $1, 4, 9, 16, 25, 36, 49, 64, 81$.
Total number of favourable outcomes $=9$.
We know that,
Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$
Therefore,
Probability that the disc bears a perfect square number $=\frac{9}{90}$
$=\frac{1}{10}$
The probability that it bears a perfect square number is $\frac{1}{10}$.
(iii) Numbers divisible by 5 from 1 to 90 are $5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90$.
Total number of favourable outcomes $=18$.
We know that,
Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$
Therefore,
Probability that the disc bears a number divisible by 5 $=\frac{18}{90}$
$=\frac{1}{5}$
The probability that it bears a number divisible by 5 is $\frac{1}{5}$.
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