Q18(ii):
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 (ii) a perfect square number

Solution :

Given: A box contains 90 discs, numbered from 1 to 90. One disc is drawn at random.

To Find: The probability that the drawn disc bears a perfect square number.

Step 1: Determine the Total Number of Possible Outcomes

Let $S$ be the sample space of all possible outcomes. Since the discs are numbered from 1 to 90, the total number of discs is 90.

Total number of outcomes, $n(S) = 90$.

Step 2: Identify the Favorable Outcomes

Let $E$ be the event of drawing a disc that bears a perfect square number. We list all perfect square numbers between 1 and 90 inclusive:

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

$5^2 = 25$

$6^2 = 36$

$7^2 = 49$

$8^2 = 64$

$9^2 = 81$

$10^2 = 100$ (Since $100 > 90$, this is excluded.)

The set of favorable outcomes is $E = \{1, 4, 9, 16, 25, 36, 49, 64, 81\}$.

Step 3: Count the Number of Favorable Outcomes

Counting the elements in set $E$:

Number of favorable outcomes, $n(E) = 9$.

Step 4: Apply the Probability Formula

The probability of an event $E$ is given by the ratio of the number of favorable outcomes to the total number of possible outcomes:

$P(E) = \frac{n(E)}{n(S)}$ [Definition of Probability for equally likely outcomes]

$P(E) = \frac{9}{90}$

Step 5: Simplify the Fraction

$P(E) = \frac{9 \div 9}{90 \div 9}$

$P(E) = \frac{1}{10}$

Final Answer: The probability that the drawn disc bears a perfect square number is $\frac{1}{10}$ or $0.1$.

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