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. $\dfrac{81}{10^8}$

Hint:

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

Answer 1

The Correct Option is B

To find the probability of getting at least 7 defective bulbs when selecting 8 bulbs one by one with replacement from a lot containing 10 defective and 90 non-defective bulbs, we can model this situation using a binomial distribution.

The probability of selecting a defective bulb in any one trial is \(\frac{10}{100} = 0.1\), since there are 10 defective bulbs out of 100. Since the selection is with replacement, this probability remains the same for each draw.

The probability that a binomial random variable \(X\) with parameters \(n = 8\) (number of trials) and \(p = 0.1\) (probability of success on each trial) takes on a particular value is given by:

\(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\)

We need to calculate the probability of getting at least 7 defective bulbs, i.e., \(P(X \geq 7)\), which is given by:

\(P(X \geq 7) = P(X=7) + P(X=8)\)

First, calculate \(P(X=7)\):

\(P(X=7) = \binom{8}{7} (0.1)^7 (0.9)^1 = 8 \times (10^{-7}) \times 0.9 = 7.2 \times 10^{-7}\)

Next, calculate \(P(X=8)\):

\(P(X=8) = \binom{8}{8} (0.1)^8 (0.9)^0 = (10^{-8}) = 10^{-8}\)

Adding these two probabilities:

\(P(X \geq 7) = 7.2 \times 10^{-7} + 10^{-8} = 72 \times 10^{-8} + 1 \times 10^{-8} = 73 \times 10^{-8}\)

Thus, the probability of getting at least 7 defective bulbs is \(\frac{73}{10^8}\), which matches the given correct answer.

The correct answer is:

\(\frac{73}{10^8}\)