A manufacturer of liquid crystal displays (LCDs) is studying their production lines. The probability of sampling a defective LCD is 0.1. A sample of 5 LCDs is taken. You may assume that an LCD being defective is independent of any of the others being defective.
What is the probability that all 5 LCDs are NOT defective?
The probability of an LCD being defective is 0.1, which means the probability of an LCD not being defective is 1 - 0.1 = 0.9.
Since the LCDs being defective are independent events, the probability of all 5 LCDs not being defective is the product of the probabilities of each LCD not being defective:
P(not defective, not defective, not defective, not defective, not defective) = 0.9 * 0.9 * 0.9 * 0.9 * 0.9
= (0.9) ^ 5
= 0.59049
So, the probability that all 5 LCDs are not defective is approximately 0.59049.
To find the probability that all five LCDs are not defective, we need to find the probability of each individual LCD not being defective and then multiply them together.
Given that the probability of sampling a defective LCD is 0.1, the probability of not sampling a defective LCD is 1 - 0.1, which is 0.9.
Since the samples are independent, the probability of all five LCDs not being defective can be found by multiplying the individual probabilities together:
Probability of not sampling a defective LCD = 0.9
Probability that all 5 LCDs are not defective = (0.9)*(0.9)*(0.9)*(0.9)*(0.9) = 0.9^5 = 0.59049
Therefore, the probability that all 5 LCDs are not defective is approximately 0.59049.
To find the probability that all 5 LCDs are not defective, we can assume that the probability of any particular LCD being defective is 0.1 (as given in the problem statement).
Since the events of each LCD being defective or not defective are independent, we can use the multiplication rule of probability to find the probability that all 5 LCDs are not defective.
The probability that any particular LCD is not defective is given by (1 - probability of being defective), which is (1 - 0.1) = 0.9.
Using the multiplication rule of probability, we can multiply the probabilities of all 5 LCDs being not defective together to get the probability that none of them are defective:
P(All 5 LCDs not defective) = P(LCD1 not defective) * P(LCD2 not defective) * P(LCD3 not defective) * P(LCD4 not defective) * P(LCD5 not defective)
= 0.9 * 0.9 * 0.9 * 0.9 * 0.9
= 0.9^5
= 0.59049
Therefore, the probability that all 5 LCDs are not defective is approximately 0.59049, or 59.049%.