Question

A lot of 100 bulbs contains 10 defective bulbs. Five bulbs are selected at random from the lot and are sent to the retail store. Then the probability that the store will receive at most one defective bulb is:

• A. \( \frac{7}{5} \left( \frac{9}{10} \right)^4 \)
• B. \( \frac{7}{5} \left( \frac{9}{10} \right)^5 \)
• C. \( \frac{6}{5} \left( \frac{9}{10} \right)^4 \)
• D. \( \frac{6}{5} \left( \frac{9}{10} \right)^5 \)

Hint:

For probability problems involving selection without replacement, use the hypergeometric distribution formula to calculate the probabilities of selecting various numbers of a specific type of object.

Answer 1

The Correct Option is A

We are given the following: - Total number of bulbs = 100
- Number of defective bulbs = 10
- Number of non-defective bulbs = 90
- Number of bulbs selected = 5
We need to calculate the probability that at most one defective bulb is selected. Step 1: Probability of selecting 0 defective bulbs To select 0 defective bulbs, we must select all 5 bulbs from the 90 non-defective ones. The number of ways to select 5 non-defective bulbs from 90 is: \[ \binom{90}{5} = \frac{90!}{5!(90-5)!} \] The total number of ways to select 5 bulbs from 100 is: \[ \binom{100}{5} = \frac{100!}{5!(100-5)!} \] Thus, the probability of selecting 0 defective bulbs is: \[ P(X = 0) = \frac{\binom{90}{5}}{\binom{100}{5}} = \frac{(90)(89)(88)(87)(86)}{(100)(99)(98)(97)(96)} \approx 0.5837 \] Step 2: Probability of selecting 1 defective bulb To select 1 defective bulb, we need to select 1 defective bulb from the 10 defective bulbs, and 4 non-defective bulbs from the 90 non-defective bulbs. The number of ways to do this is: \[ \binom{10}{1} \binom{90}{4} = 10 \times \frac{90!}{4!(90-4)!} \] Thus, the probability of selecting 1 defective bulb is: \[ P(X = 1) = \frac{\binom{10}{1} \binom{90}{4}}{\binom{100}{5}} \approx 0.31 \] Step 3: Total probability of selecting at most 1 defective bulb The total probability of selecting at most 1 defective bulb is the sum of the probabilities for selecting 0 and 1 defective bulb: \[ P(X \leq 1) = P(X = 0) + P(X = 1) = 0.5837 + 0.31 = 0.8937 \] Step 4: Use of Hypergeometric Distribution This problem can be solved using the hypergeometric distribution. The formula for the hypergeometric distribution is: \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \] Where:
- \( N \) = total number of bulbs = 100
- \( K \) = number of defective bulbs = 10
- \( n \) = number of bulbs selected = 5
- \( k \) = number of defective bulbs selected
Thus, the probability \( P(X \leq 1) \) can be written as: \[ P(X = 0) = \frac{\binom{10}{0} \binom{90}{5}}{\binom{100}{5}}, \quad P(X = 1) = \frac{\binom{10}{1} \binom{90}{4}}{\binom{100}{5}}. \] So: \[ P(X \leq 1) = P(X = 0) + P(X = 1) \approx 0.8937. \] This matches the option: \[ \boxed{\frac{7}{5} \left( \frac{9}{10} \right)^4}. \]