Question

There are 10 defective and 90 non-defective balls in a bag. 8 balls are taken one by one with replacement. Find the probability that at least 7 defective balls are selected.

• A. $\dfrac{73}{10^{8}}$
• B. $\dfrac{37}{10^{8}}$
• C. $\dfrac{105}{10^{8}}$
• D. $\dfrac{11}{10^{8}}$

Hint:

For repeated trials with replacement, always use the binomial distribution since each trial is independent.

Answer 1

The Correct Option is A

Step 1: Identify the probability of selecting a defective ball.
Since there are 10 defective balls out of 100 balls, \[ P(\text{defective}) = \frac{10}{100} = \frac{1}{10} \] and \[ P(\text{non-defective}) = \frac{9}{10} \] Step 2: Use the binomial probability formula.
Balls are drawn with replacement, so trials are independent.
We need the probability of getting at least 7 defective balls out of 8: \[ P(X \ge 7) = P(X=7) + P(X=8) \] Step 3: Compute each probability.
\[ P(X=7) = \binom{8}{7}\left(\frac{1}{10}\right)^7\left(\frac{9}{10}\right) = \frac{72}{10^8} \] \[ P(X=8) = \binom{8}{8}\left(\frac{1}{10}\right)^8 = \frac{1}{10^8} \] Step 4: Add the probabilities.
\[ P(X \ge 7) = \frac{72}{10^8} + \frac{1}{10^8} = \frac{73}{10^8} \]