A batch of one hundred bulbs is inspected by testing four randomly cho...
Question: A batch of one hundred bulbs is inspected by testing four randomly chosen bulbs. The batch is rejected if even one of the bulbs is defective. A batch typically has five defective bulbs. The probability that the current batch is accepted is _________ (Important - Enter only the numerical value in the answer).
Answer:
To find the probability that the current batch is accepted, we need to calculate the probability that all four randomly chosen bulbs are non-defective.
Step 1: Finding the probability of selecting a non-defective bulb in the first attempt
Since there are 100 bulbs in the batch and five of them are defective, the probability of selecting a non-defective bulb in the first attempt is:
P(non-defective on first attempt) = (100 - 5) / 100 = 95 / 100 = 0.95
Step 2: Finding the probability of selecting a non-defective bulb in the second attempt
After the first non-defective bulb has been selected, there are 99 bulbs remaining in the batch, out of which four are defective. So, the probability of selecting a non-defective bulb in the second attempt is:
P(non-defective on second attempt) = (99 - 4) / 99 = 95 / 99 ≈ 0.9596
Step 3: Finding the probability of selecting a non-defective bulb in the third attempt
After the second non-defective bulb has been selected, there are 98 bulbs remaining in the batch, out of which three are defective. So, the probability of selecting a non-defective bulb in the third attempt is:
P(non-defective on third attempt) = (98 - 3) / 98 = 95 / 98 ≈ 0.9694
Step 4: Finding the probability of selecting a non-defective bulb in the fourth attempt
After the third non-defective bulb has been selected, there are 97 bulbs remaining in the batch, out of which two are defective. So, the probability of selecting a non-defective bulb in the fourth attempt is:
P(non-defective on fourth attempt) = (97 - 2) / 97 = 95 / 97 ≈ 0.9794
Step 5: Calculating the probability of the current batch being accepted
To calculate the probability that all four randomly chosen bulbs are non-defective, we multiply the probabilities from each step together:
P(batch accepted) = P(non-defective on first attempt) * P(non-defective on second attempt) * P(non-defective on third attempt) * P(non-defective on fourth attempt)
P(batch accepted) ≈ 0.95 * 0.9596 * 0.9694 * 0.9794 ≈ 0.8145
Therefore, the probability that the current batch is accepted is approximately 0.8145.