Provided information:
- Total bulbs: 100
- Defective bulbs: 10
- Non-defective bulbs: 90
- Bulbs selected: 5
Goal: Calculate the probability of selecting at most one defective bulb.
Step 1: Probability of selecting 0 defective bulbs
This requires selecting all 5 bulbs from the 90 non-defective ones. The number of ways to do this is:
\[
\binom{90}{5} = \frac{90!}{5!(90-5)!}
\]
The total number of ways to select 5 bulbs from 100 is:
\[
\binom{100}{5} = \frac{100!}{5!(100-5)!}
\]
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
This involves selecting 1 defective bulb from 10 and 4 non-defective bulbs from 90. The number of ways is:
\[
\binom{10}{1} \binom{90}{4} = 10 \times \frac{90!}{4!(90-4)!}
\]
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
This is the sum of probabilities from Step 1 and Step 2:
\[
P(X \leq 1) = P(X = 0) + P(X = 1) = 0.5837 + 0.31 = 0.8937
\]
Step 4: Application of Hypergeometric Distribution
This problem is solvable using the hypergeometric distribution formula:
\[
P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}
\]
Where:
- \( N \) = 100 (total bulbs)
- \( K \) = 10 (defective bulbs)
- \( n \) = 5 (bulbs selected)
- \( k \) = number of defective bulbs selected
Thus, \( P(X \leq 1) \) is calculated as:
\[
P(X = 0) = \frac{\binom{10}{0} \binom{90}{5}}{\binom{100}{5}}, \quad P(X = 1) = \frac{\binom{10}{1} \binom{90}{4}}{\binom{100}{5}}.
\]
Therefore:
\[
P(X \leq 1) = P(X = 0) + P(X = 1) \approx 0.8937.
\]
This result aligns with the option:
\[
\boxed{\frac{7}{5} \left( \frac{9}{10} \right)^4}.
\]