Question

From a lot containing $10$ defective and $90$ non-defective bulbs, $8$ bulbs are selected one by one with replacement. Then the probability of getting at least $7$ defective bulbs is

• A. $\dfrac{67}{10^8}$
• B. $\dfrac{73}{10^8}$
• C. $\dfrac{7}{10^7}$
• D. $\dfrac{81}{10^8}$
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Hint:

When trials are independent and probabilities remain constant, use the binomial distribution directly.

Answer 1

The Correct Option is B

Step 1: Identifying the probability model.
Since bulbs are selected with replacement, the probability of selecting a defective bulb remains constant for each trial.
\[ P(\text{defective})=\frac{10}{100}=\frac{1}{10} \] Thus, the number of defective bulbs selected follows a binomial distribution with \[ n=8,\quad p=\frac{1}{10} \] Step 2: Writing the required probability.
“At least $7$ defective bulbs” means \[ P(X\ge7)=P(X=7)+P(X=8) \] Step 3: Calculating $P(X=7)$.
\[ P(X=7)=\binom{8}{7}\left(\frac{1}{10}\right)^7\left(\frac{9}{10}\right) \] \[ =\frac{72}{10^8} \] Step 4: Calculating $P(X=8)$.
\[ P(X=8)=\binom{8}{8}\left(\frac{1}{10}\right)^8 =\frac{1}{10^8} \] Step 5: Final computation.
\[ P(X\ge7)=\frac{72}{10^8}+\frac{1}{10^8} =\frac{73}{10^8} \]