Question

A lot of 100 bulbs contains 10 defective bulbs. Five bulbs selected at random from the lot and sent to retain store, then the probability that the store will receive at most one defective bulb is

• A. 0.59049
• B. 0.91854
• C. 0.6561
• D. 0.32805

Hint:

To save time during calculations without a calculator, you can factor out common terms! $P(X=0) + P(X=1) = (0.9)^5 + 5(0.1)(0.9)^4 = (0.9)^4 \times [0.9 + 0.5] = 0.6561 \times 1.4 = 0.91854$. Adding inside the bracket first means fewer large decimals to multiply at the end.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are selecting a small random sample (5 bulbs) from a larger batch (100 bulbs) that has a known defect rate. We need to find the overall probability of getting "at most one" defective bulb in our selected sample.

Step 2: Key Formula or Approach:
Because the sample size ($n=5$) is very small compared to the total population size ($N=100$), we can reliably approximate the probability using the Binomial Distribution formula.
The probability mass function for a binomial distribution is:
$$P(X = k) = \binom{n}{k} p^k q^{n-k}$$ The phrase "at most one" implies $X \le 1$, so we must calculate and sum the probabilities of exactly zero defective bulbs and exactly one defective bulb: $P(X = 0) + P(X = 1)$.

Step 3: Detailed Explanation:
First, find the probability of a single bulb being defective ($p$) and non-defective ($q$):
$$p = \frac{10}{100} = 0.1$$ $$q = 1 - p = 1 - 0.1 = 0.9$$ Number of trials (bulbs selected) is $n = 5$.
Calculate the probability of getting exactly zero defective bulbs ($X = 0$):
$$P(X = 0) = \binom{5}{0} (0.1)^0 (0.9)^5 = 1 \times 1 \times (0.9)^5$$ $$P(X = 0) = 0.59049$$ Calculate the probability of getting exactly one defective bulb ($X = 1$):
$$P(X = 1) = \binom{5}{1} (0.1)^1 (0.9)^4 = 5 \times 0.1 \times (0.9)^4$$ $$P(X = 1) = 0.5 \times 0.6561 = 0.32805$$ Add them together to find the total cumulative probability:
$$P(X \le 1) = P(X = 0) + P(X = 1)$$ $$P(X \le 1) = 0.59049 + 0.32805 = 0.91854$$

Step 4: Final Answer:
The probability of receiving at most one defective bulb is $0.91854$, corresponding to option (B).