In a large lot of electric light bulbs,5% of the bulbs are defective.calculate the probability that a random sample of 20 will contain at most 2 defective bulbs? Try ;->Pr=nCx p^xq^n-x .pr(2D)=20C2(0.05)^2(0.95)^18;->=2.2/ANS

To calculate the probability that a random sample of 20 light bulbs will contain at most 2 defective bulbs, you can use the formula:

P(x ≤ 2) = nCx * p^x * q^(n-x)

Here, n represents the sample size (20 bulbs), x represents the number of defective bulbs (0, 1, or 2), p represents the probability of a bulb being defective (5% or 0.05), and q represents the probability of a bulb not being defective (95% or 0.95).

Now let's substitute the values into the formula to calculate the probability:

P(0 ≤ x ≤ 2) = 20C0 * (0.05)^0 * (0.95)^(20-0)
+ 20C1 * (0.05)^1 * (0.95)^(20-1)
+ 20C2 * (0.05)^2 * (0.95)^(20-2)

Calculating these values:

P(0) = 20C0 * (0.05)^0 * (0.95)^20 = 1 * 1 * 0.3584859224 = 0.3585 (rounded)
P(1) = 20C1 * (0.05)^1 * (0.95)^19 = 20 * 0.05 * 0.3783808006 = 0.3784 (rounded)
P(2) = 20C2 * (0.05)^2 * (0.95)^18 = 190 * 0.0025 * 0.4032879206 = 0.1924 (rounded)

Now, add these probabilities together:

P(0 ≤ x ≤ 2) = P(0) + P(1) + P(2) = 0.3585 + 0.3784 + 0.1924 = 0.9293 (rounded)

Therefore, the probability that a random sample of 20 light bulbs will contain at most 2 defective bulbs is approximately 0.9293.

To calculate the probability that a random sample of 20 light bulbs will contain at most 2 defective bulbs, you can use the binomial probability formula. The formula is as follows:

Pr = nCx * p^x * q^(n-x)

Where:
Pr is the probability
n is the total number of trials or light bulbs in this case (20)
x is the number of successes (defective bulbs)
p is the probability of success (5% or 0.05 in decimal form)
q is the probability of failure (100% - p or 95% or 0.95 in decimal form)
nCx is the binomial coefficient, which represents the number of ways to choose x successes from n trials.

Now, let's calculate the probability:

Pr(2D) = 20C2 * (0.05)^2 * (0.95)^18

Pr(2D) = 20! / (2! * (20-2)!) * (0.05)^2 * (0.95)^18

Pr(2D) = 20 * 19 / (2 * 1) * (0.05)^2 * (0.95)^18

Pr(2D) = 190 / 2 * (0.0025) * (0.3584859224)

Pr(2D) ≈ 0.0178776326

So, the probability that a random sample of 20 light bulbs will contain at most 2 defective bulbs is approximately 0.0179 or 1.79%.