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

To solve this problem, we need to find the probability that at least 7 defective balls are selected from a bag containing 10 defective and 90 non-defective balls when 8 balls are drawn one by one with replacement.

1. First, determine the probability of selecting a defective ball in one draw. \(P(\text{defective}) = \frac{10}{100} = \frac{1}{10}\)
2. Since the draws are with replacement, each draw is independent. Therefore, the probability of selecting a non-defective ball in one draw is: \(P(\text{non-defective}) = \frac{90}{100} = \frac{9}{10}\)
3. We need to find the probability of getting at least 7 defective balls out of 8 draws. This can be expressed as: \(P(\text{at least 7 defective}) = P(\text{7 defective}) + P(\text{8 defective})\)
4. Calculate the probability of exactly 7 defective balls: \(P(\text{7 defective}) = \binom{8}{7} \left( \frac{1}{10} \right)^7 \left( \frac{9}{10} \right)^1\) Simplifying: \(= 8 \cdot \frac{1}{10^7} \cdot \frac{9}{10} = \frac{72}{10^8}\)
5. Calculate the probability of exactly 8 defective balls: \(P(\text{8 defective}) = \binom{8}{8} \left( \frac{1}{10} \right)^8 \left( \frac{9}{10} \right)^0\) Simplifying: \(= 1 \cdot \frac{1}{10^8} = \frac{1}{10^8}\)
6. Now, add the probabilities of these two events: \(P(\text{at least 7 defective}) = \frac{72}{10^8} + \frac{1}{10^8} = \frac{73}{10^8}\)

Therefore, the probability that at least 7 defective balls are selected is \(\frac{73}{10^8}\).