Questions: A factory produces light bulbs with a defect rate of 10%. If a quality control inspector randomly selects 10 light bulbs, what is the probability that: a) Exactly 4 of the light bulbs are defective? b) Fewer than 6 of the light bulbs are defective?

Solution Steps

To find the probability that exactly 4 out of 10 light bulbs are defective, we use the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

where:

• \( n = 10 \) (total number of bulbs),
• \( x = 4 \) (number of defective bulbs),
• \( p = 0.1 \) (probability of a bulb being defective),
• \( q = 1 - p = 0.9 \) (probability of a bulb being non-defective).

Calculating this gives:

\[ P(X = 4) = \binom{10}{4} \cdot (0.1)^4 \cdot (0.9)^{6} \approx 0.0112 \]

Thus, the probability of exactly 4 defective bulbs is \( 0.0112 \).

To find the probability that fewer than 6 bulbs are defective, we sum the probabilities of having 0, 1, 2, 3, 4, and 5 defective bulbs:

\[ P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) \]

Calculating each probability:

• \( P(X = 0) \approx 0.3487 \)
• \( P(X = 1) \approx 0.3874 \)
• \( P(X = 2) \approx 0.1937 \)
• \( P(X = 3) \approx 0.0574 \)
• \( P(X = 4) \approx 0.0112 \)
• \( P(X = 5) \approx 0.0015 \)

Summing these probabilities:

\[ P(X < 6) \approx 0.3487 + 0.3874 + 0.1937 + 0.0574 + 0.0112 + 0.0015 \approx 0.9999 \]

Thus, the probability of fewer than 6 defective bulbs is approximately \( 0.9999 \).

Final Answer