To find the probability that more than 3 LCDs are NOT defective, we can find the probability that 3 or fewer LCDs ARE defective and subtract it from 1.
The probability of an LCD being defective is 0.1. Therefore, the probability of an LCD NOT being defective is 1 - 0.1 = 0.9.
Since the probability of an LCD not being defective is 0.9, the probability of 3 or fewer LCDs being defective can be found using the binomial probability formula:
P(X ≤ 3) = C(5, 0) * (0.9)^5 + C(5, 1) * (0.9)^4 * (0.1)^1 + C(5, 2) * (0.9)^3 * (0.1)^2 + C(5, 3) * (0.9)^2 * (0.1)^3
where C(n, r) represents the combination of choosing r items from a set of n items.
Using this formula, we can calculate P(X ≤ 3) = 0.0324.
Finally, subtracting P(X ≤ 3) from 1, we have:
P(X > 3) = 1 - P(X ≤ 3) = 1 - 0.0324 = 0.9676.
Therefore, the probability that more than 3 LCDs are NOT defective is 0.9676 or 96.76%.
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 more than 3 LCDs are NOT defective?
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