Problem 24 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

In a box of 10 electric bulbs, 2 are defective. Two bulbs are selected at random one after the other from the box. The first bulb after selection being put back in the box before making the second selection. The probability that both the bulbs are without defect is (a) \(9 / 25\) (c) \(4 / 5\) (b) \(16 / 25\) (d) \(8 / 25\)

Step-by-Step Solution

Answer

\( \frac{16}{25} \)

Step 1: Calculate Probability of First Selection

Since there are 10 bulbs and 2 are defective, there are 8 non-defective bulbs. The probability of selecting a non-defective bulb on the first try is calculated as: \( \text{Probability of selecting a non-defective bulb first} = \frac{8}{10} = \frac{4}{5} \).

Step 2: Calculate Probability of Second Selection

Since the first bulb is replaced back in the box, the situation resets and there are still 10 bulbs including all non-defective ones. The probability of selecting a non-defective bulb again is: \( \text{Probability of selecting a non-defective bulb second} = \frac{8}{10} = \frac{4}{5} \).

Step 3: Combine Probabilities for Both Selections

The processes are independent, so multiply the probabilities of each event. Therefore, the probability that both bulbs are without defect is: \[ \left( \frac{4}{5} \right) \cdot \left( \frac{4}{5} \right) = \frac{16}{25} \].

Key Concepts

Independent events

In probability, two events are considered independent if the occurrence or outcome of one event does not affect the other. This means that whatever happens with the first event has no influence on the likelihood of the second event occurring. In the context of our bulb selection problem, the independence comes from the fact that after the first selection, the chosen bulb is replaced back into the box, resetting the conditions for the second draw.

• First selection: Choosing a bulb.
• Replacement: Putting the bulb back into the box.
• Second selection: Choosing a bulb again.

Because the bulb is replaced, the second selection starts with the same initial conditions as the first, ensuring that the two draws are independent events. This resets the total number of bulbs and the number of non-defective bulbs, making it as though the first draw never happened, and thus the probability remains constant.

Replacement in probability

The concept of replacement in probability is essential when dealing with multiple events and ensuring they remain independent. Replacement refers to the action of putting an item back into the pool of choices after it has been drawn. This keeps the total number of items constant across multiple events.

• Ensures independence: Replacement allows each selection to be uninfluenced by the previous one.
• Consistent statistics: Keeps the number of defective and non-defective bulbs unchanged for every draw.

In our bulb example, when the first non-defective bulb is returned to the box, the conditions for the second selection are identical to the first. This means each draw's probability calculation is the same, as the scenario resets after each selection.

Non-defective selection

Understanding the selection of non-defective bulbs is crucial in this probability problem. Initially, there are 10 electric bulbs in the box, out of which 8 are non-defective. So, the probability of picking a non-defective bulb is formulated by comparing the number of successful outcomes to the total number of possible outcomes.

• Total bulbs: 10
• Non-defective bulbs: 8

The probability of drawing a non-defective bulb first is given by: \(\frac{8}{10} = \frac{4}{5}\). Since the selection involves replacement, this probability remains constant for subsequent draws, allowing us to apply the same formula every time a draw is made.

Event multiplication

Event multiplication is a fundamental concept in probability used to find the likelihood of multiple independent events happening in sequence. For this, the probabilities of individual events are multiplied. In the case of our problem, both draws of the non-defective bulbs are treated as independent events due to replacement.

• Probability of first non-defective bulb: \(\frac{4}{5}\)
• Probability of second non-defective bulb: \(\frac{4}{5}\)
• Total probability for both: \(\left( \frac{4}{5} \right) \cdot \left( \frac{4}{5} \right)\)

The calculated probability using event multiplication yields \(\frac{16}{25}\), which answers the question of both bulbs being non-defective in consecutive selections. This technique of multiplying probabilities is key for sequential independent events to determine a combined probability.