A batch of one hundred bulbs is inspected by testing four randomly cho...
Probability for one bulb to be non defective is

∴ Probabilities that none of the bulbs is defectives

Binomial Problem with n = 4 and p(not defective) = 0.95
P(x = 4 non defective) = 0.95 x 4 = 0.8145
A batch of one hundred bulbs is inspected by testing four randomly cho...
To solve this problem, we can use the concept of probability. Let's break down the problem step by step:
1. Finding the probability of selecting a defective bulb:
Since there are 100 bulbs in the batch and 5 of them are defective, the probability of selecting a defective bulb is 5/100 = 1/20.
2. Finding the probability of selecting a non-defective bulb:
The probability of selecting a non-defective bulb is the complement of the probability of selecting a defective bulb. So, it is 1 - 1/20 = 19/20.
3. Finding the probability of all four selected bulbs being non-defective:
Since we are testing four bulbs, and each bulb is selected randomly and independently, the probability of all four bulbs being non-defective is (19/20) * (19/20) * (19/20) * (19/20) = (19/20)^4.
4. Finding the probability of at least one bulb being defective:
The probability of at least one bulb being defective is the complement of the probability of all four bulbs being non-defective. So, it is 1 - (19/20)^4.
5. Calculating the final probability:
The final probability of the batch being accepted is the probability of at least one bulb being defective. So, it is 1 - (19/20)^4.
Calculating this probability gives us the answer as approximately 0.8145, which corresponds to option C.
Therefore, the correct answer is option C) 0.8145.