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Q17(ii):
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?

Solution :

Given:

From the context of the previous problem (Question 17), a lot of 20 bulbs contains 4 defective bulbs. In part (i), one bulb was drawn and it was found to be non-defective. This bulb is not replaced.

To Find:

The probability that a bulb drawn at random from the remaining bulbs is not defective.

Step 1: Analyzing the initial state of the lot.

Total number of bulbs = $20$

Number of defective bulbs = $4$

Number of non-defective bulbs = $20 - 4 = 16$

Step 2: Updating the state after the first draw.

In part (i), one non-defective bulb was drawn and not replaced. We update the counts as follows:

New total number of bulbs = $20 - 1 = 19$

Since the bulb drawn was non-defective, the number of defective bulbs remains unchanged: $4$

New number of non-defective bulbs = $16 - 1 = 15$

Step 3: Defining the Probability Formula.

The probability $P(E)$ of an event $E$ is defined as:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Step 4: Calculating the probability of drawing a non-defective bulb.

Let $E$ be the event of drawing a non-defective bulb from the remaining lot.

Number of favorable outcomes (non-defective bulbs remaining) = $15$

Total number of possible outcomes (total bulbs remaining) = $19$

$P(E) = \frac{15}{19}$

[Since the number of non-defective bulbs is 15 and the total number of bulbs is 19]

Final Answer:

Final Answer: The probability that the bulb drawn is not defective is $\frac{15}{19}$.


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