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: Treat it as binomial.
Each bulb is defective with $p = 0.1$ and good with $q = 0.9$, and we pick $n = 5$.

Step 2: Add the zero and one cases.
$$P(X=0) = (0.9)^5 = 0.59049$$
$$P(X=1) = 5(0.1)(0.9)^4 = 0.32805$$

Step 3: Sum for "at most one".
$$0.59049 + 0.32805 = 0.91854$$
\[ \boxed{0.91854} \]