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Q21(ii):
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (ii) She will not buy it ?

Solution :

Given:

Total number of ball pens in the lot = $144$

Number of defective pens = $20$

Condition for buying: Nuri buys the pen if it is good; she does not buy the pen if it is defective.

To Find:

The probability that Nuri will not buy the pen.

Step 1: Defining the Sample Space

Let $S$ be the sample space representing the total number of pens available in the lot. The total number of outcomes is given by the total number of pens.

$n(S) = 144$

Step 2: Defining the Event

Let $E$ be the event that Nuri will not buy the pen. According to the problem statement, Nuri will not buy the pen if it is defective.

Therefore, the number of favorable outcomes for event $E$ is equal to the number of defective pens.

$n(E) = 20$

Step 3: Applying the Probability Formula

The probability of an event $P(E)$ is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes in the sample space.

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

Substituting the known values:

$P(E) = \frac{20}{144}$

Step 4: Simplifying the Fraction

To simplify $\frac{20}{144}$, we find the greatest common divisor (GCD) of $20$ and $144$.

Prime factorization of $20$: $2^2 \times 5$

Prime factorization of $144$: $2^4 \times 3^2$

The common factor is $2^2 = 4$.

Dividing both the numerator and the denominator by $4$:

$P(E) = \frac{20 \div 4}{144 \div 4}$

$P(E) = \frac{5}{36}$

Final Answer: The probability that Nuri will not buy the pen is $\frac{5}{36}$.


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