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 exactly 4 LCDs are NOT defective?
To find the probability that exactly 4 LCDs are not defective, we need to find the probability of selecting 4 non-defective LCDs out of 5 samples.
The probability of selecting a non-defective LCD is 1 - 0.1 = 0.9.
The probability of selecting exactly 4 non-defective LCDs out of 5 samples can be found using the binomial probability formula:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
n is the number of trials (5 in this case)
k is the number of successful outcomes (4 in this case)
p is the probability of a successful outcome (0.9 in this case)
P(X=4) = (5 choose 4) * 0.9^4 * (1-0.9)^(5-4)
= (5 choose 4) * 0.9^4 * 0.1^1
= 5 * 0.6561 * 0.1
= 0.32805
Therefore, the probability that exactly 4 LCDs are not defective is approximately 0.328.
To calculate the probability that exactly 4 LCDs are not defective, we need to consider the probability of each individual LCD being not defective.
Given that the probability of sampling a defective LCD is 0.1, the probability of sampling a non-defective LCD is 1 - 0.1 = 0.9.
Since there are 5 LCDs in the sample, we need to calculate the probability of exactly 4 of them being not defective.
Using the binomial probability formula, the probability of exactly 4 successes (not defective LCDs) out of 5 trials is:
P(X = 4) = (5 Choose 4) * (0.9)^4 * (0.1)^1
Using the binomial coefficient formula:
(5 Choose 4) = 5! / (4!(5-4)!) = 5
Plugging in the values:
P(X = 4) = 5 * (0.9)^4 * (0.1)^1
P(X = 4) = 0.32805
Therefore, the probability that exactly 4 LCDs are not defective is approximately 0.32805.
To find the probability that exactly 4 LCDs are not defective, we can use the binomial probability formula. The formula is given by:
P(X = k) = (n C k) * p^k * q^(n-k)
Where:
P(X = k) is the probability that exactly k LCDs are not defective,
n is the sample size, which is 5 in this case,
k is the number of LCDs that are not defective,
p is the probability that an LCD is defective, which is 0.1 in this case,
q is the probability that an LCD is not defective, which is 1 - p = 1 - 0.1 = 0.9,
(n C k) is the number of combinations of n items taken k at a time, which can be calculated as n! / (k! * (n-k)!).
Plugging in the values into the formula:
P(X = 4) = (5 C 4) * (0.1)^4 * (0.9)^(5-4)
Calculating the combinations:
(5 C 4) = 5! / (4! * (5-4)!) = 5
Now, we can substitute the values into the formula:
P(X = 4) = 5 * (0.1)^4 * (0.9)^1
Simplifying the equation:
P(X = 4) = 5 * 0.0001 * 0.9
P(X = 4) = 0.00045
Therefore, the probability that exactly 4 LCDs are not defective is 0.00045 or 0.045%.