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: Set up the chance for one bulb.
Out of $100$ bulbs, $10$ are defective, so the chance a bulb is defective is $p=\frac{10}{100}=0.1$ and good is $q=0.9$. We pick $n=5$ bulbs.
Step 2: Understand at most one.
At most one defective means zero defective or exactly one defective. We add those two chances.
Step 3: Chance of zero defective.
$P(X=0)=q^5=(0.9)^5=0.59049$.
Step 4: Chance of exactly one defective.
$P(X=1)=\binom{5}{1}p\,q^4=5(0.1)(0.9)^4=5(0.1)(0.6561)=0.32805$.
Step 5: Add them up.
$P(X\le 1)=0.59049+0.32805=0.91854$.
Step 6: State the answer.
The store gets at most one defective bulb with probability $0.91854$, which is option (2). \[ \boxed{P(X\le 1)=0.91854} \]