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Q18(i):
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

Solution :

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

To find: The probability that the drawn disc bears a two-digit number.

Step 1: Determine the Total Number of Possible Outcomes

The discs are numbered from 1 to 90. Therefore, the total number of discs in the box is 90.

Let $S$ be the sample space of all possible outcomes.

$n(S) = 90$

Step 2: Identify the Favorable Outcomes

We need to find the number of discs that bear a two-digit number. The numbers on the discs range from 1 to 90.

The numbers from 1 to 9 are single-digit numbers. There are 9 such numbers ($1, 2, 3, 4, 5, 6, 7, 8, 9$).

The numbers from 10 to 90 are two-digit numbers.

To find the count of numbers from 10 to 90 inclusive, we use the formula for the number of terms in an arithmetic progression: $n = (\text{Last term} - \text{First term}) + 1$.

Number of two-digit discs = $(90 - 10) + 1$

Number of two-digit discs = $80 + 1 = 81$

Let $E$ be the event of drawing a disc with a two-digit number.

$n(E) = 81$

Step 3: 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:

$P(E) = \frac{n(E)}{n(S)}$

[Substituting the values identified in Step 1 and Step 2]

$P(E) = \frac{81}{90}$

Step 4: Simplify the Fraction

To simplify $\frac{81}{90}$, we find the greatest common divisor (GCD) of 81 and 90, which is 9.

$P(E) = \frac{81 \div 9}{90 \div 9}$

$P(E) = \frac{9}{10}$

[Converting to decimal form for clarity]

$P(E) = 0.9$

Final Answer: The probability that the drawn disc bears a two-digit number is $\frac{9}{10}$ or 0.9.


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