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.
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Answer 1

$\huge\boxed{\fcolorbox{white}{pink}{Answer}}$

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The total numbers of discs = 50

P(E) = (Number of favourable outcomes/ Total number of outcomes)

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(i) Total number of discs having two digit numbers = 81

(Since 1 to 9 are single digit numbers and so, total 2 digit numbers are 90 – 9 = 81)

P (bearing a two-digit number) = 81/90 = 9/10 = 0.9

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(ii) Total number of perfect square numbers = 9 (1, 4, 9, 16, 25, 36, 49, 64 and 81)

P (getting a perfect square number) = 9/90 = 1/10 = 0.1

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(iii) Total numbers which are divisible by 5 = 18 (5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90)

P (getting a number divisible by 5) = 18/90 = ⅕ = 0.2

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Answer 2

★Answer :-

• Total outcome for every part is same = 90

1.

• There are 81 two digits number from 1 to 90 so possible outcomes are 81

$\sf{\boxed{\red{Probability = \dfrac{Favourable\:Outcome}{Total\:Outcome}}}}$

$\sf{Probability = \dfrac{81}{90}}$

So , probability of getting a two digit number is $\sf\dfrac{81}{90}$

2.

• There are 9 perfect square till 90 . So, the possibile outcomes are 9.

$\sf{\boxed{\red{Probability = \dfrac{Favourable\:Outcome}{Total\:Outcome}}}}$

$\sf{Probability = \dfrac{9}{90}}$

So , probability of getting a perfect Square number is $\sf\dfrac{1}{10}$

3.

• There are 18 number between 1 and 90 which are divisible by 5 so possible outcomes are 5

$\sf{\boxed{\red{Probability = \dfrac{Favourable\:Outcome}{Total\:Outcome}}}}$

$\sf{Probability = \dfrac{5}{90}}$

$\sf{Probability = \dfrac{1}{18}}$

So , probability of getting a two digit number is $\sf\dfrac{1}{18}$