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A box contains 120 discs, which are numbered from 1 to 120. If one disc is drawn at random from the box, find the probability that: a) it bears a 2- digit number b) the number is a perfect square.
A box contains 120 discs, which are numbered from 1 to 120. If one disc is drawn at random from the box, find the probability that:
a) it bears a 2- digit number
b) the number is a perfect square.
This Question has 1 answers.
We are given a box containing $120$ discs numbered from $1$ to $120$. A disc is drawn at random, and we need to find the probability of the following events:
### (a) Probability of Drawing a Two-Digit Number
A two-digit number ranges from $10$ to $99$. The total count of such numbers is:
$99 - 10 + 1 = 90$
Since there are $120$ discs, the probability is:
$ \displaystyle P(\text{two-digit number}) = \frac{90}{120} = \frac{3}{4} $
### (b) Probability of Drawing a Perfect Square
Perfect squares between $1$ and $120$ are:
$1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121$
Since $121$ is out of range, the valid perfect squares are:
$1, 4, 9, 16, 25, 36, 49, 64, 81, 100$
Total count = $10$
The probability is:
$ \displaystyle P(\text{perfect square}) = \frac{10}{120} = \frac{1}{12} $
### Final Answer:
(a) $ \frac{3}{4} $
(b) $ \frac{1}{12} $
### (a) Probability of Drawing a Two-Digit Number
A two-digit number ranges from $10$ to $99$. The total count of such numbers is:
$99 - 10 + 1 = 90$
Since there are $120$ discs, the probability is:
$ \displaystyle P(\text{two-digit number}) = \frac{90}{120} = \frac{3}{4} $
### (b) Probability of Drawing a Perfect Square
Perfect squares between $1$ and $120$ are:
$1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121$
Since $121$ is out of range, the valid perfect squares are:
$1, 4, 9, 16, 25, 36, 49, 64, 81, 100$
Total count = $10$
The probability is:
$ \displaystyle P(\text{perfect square}) = \frac{10}{120} = \frac{1}{12} $
### Final Answer:
(a) $ \frac{3}{4} $
(b) $ \frac{1}{12} $
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