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?
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%.
To find the probability that more than 3 LCDs are NOT defective, we need to calculate the probability of selecting 4 or 5 non-defective LCDs.
The probability of selecting a non-defective LCD is 1 - 0.1 = 0.9.
Now, we can calculate the probability of selecting 4 non-defective LCDs:
P(4 non-defective LCDs) = (0.9)^4
And the probability of selecting 5 non-defective LCDs:
P(5 non-defective LCDs) = (0.9)^5
To calculate the probability that more than 3 LCDs are NOT defective, we need to sum up these probabilities:
P(more than 3 non-defective LCDs) = P(4 non-defective LCDs) + P(5 non-defective LCDs)
P(more than 3 non-defective LCDs) = (0.9)^4 + (0.9)^5
Calculating this sum will give us the final probability.
To find the probability that more than 3 LCDs are not defective, we need to find the probability of exactly 4 LCDs not being defective and the probability of all 5 LCDs not being defective, and then add them together.
The probability of an LCD being not defective is 1 - 0.1 = 0.9.
To calculate the probability of exactly 4 LCDs not being defective, we use the binomial probability formula:
P(X = k) = (nCr) * (p^k) * ((1-p)^(n-k))
Where:
- P(X = k) is the probability of getting exactly k successes
- n is the number of trials (sample size)
- k is the number of successes (LCDs not being defective)
- p is the probability of success (probability of an LCD not being defective)
- (nCr) is the number of combinations of n items taken r at a time
In this case:
- n = 5 (sample size)
- k = 4
- p = 0.9 (probability of an LCD not being defective)
P(X = 4) = (5C4) * (0.9^4) * (0.1^1)
To calculate (5C4):
(5C4) = 5! / (4! * (5-4)!) = 5
P(X = 4) = 5 * (0.9^4) * (0.1^1) = 0.32805
Next, we need to calculate the probability of all 5 LCDs not being defective:
P(X = 5) = (5C5) * (0.9^5) * (0.1^0)
To calculate (5C5):
(5C5) = 5! / (5! * (5-5)!) = 1
P(X = 5) = 1 * (0.9^5) * (0.1^0) = 0.59049
Now, we can add the probabilities together to find the final answer:
P(more than 3 LCDs are not defective) = P(X = 4) + P(X = 5) = 0.32805 + 0.59049 = 0.91854
Therefore, the probability that more than 3 LCDs are not defective is approximately 0.91854 (or 91.854%).