In 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 detective bulbs

The outcome consists of individual answers of true/false, probability remains constant throughout, use Binomial probability.

P(n,r,p)=nCr*p^r(1-p)^(n-r)
n=size of sample
r=(exact)number of outcomes each with probability of p

At most 2 means 0,1 or 2.
The answer is P(20,0,0.05)+P(20,1,0.05)+P(20,2,0.05)

For example,
P(20,2,0.05)=20C2*0.05^2*0.95^18
=(20*19/(1*2))*0.05^2*0.95^18
=0.1887

Complete the solution by adding the two other terms.

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

The binomial probability formula is given by:
P(x) = (nCx) * p^x * q^(n-x)

Where:
P(x) is the probability of getting exactly x successes,
n is the total number of trials or samples,
x is the number of successful events we are interested in,
p is the probability of success in any single trial, and
q is the probability of failure in any single trial (q = 1 - p).

In this case, we want to calculate the probability of getting at most 2 defective bulbs, which means x can be 0, 1, or 2.

Let's calculate the probabilities step by step:

1. Probability of 0 defective bulbs (x = 0):
P(0) = (20C0) * (0.05)^0 * (0.95)^(20-0)

To calculate (20C0) - the number of combinations of choosing 0 defective bulbs out of 20, we can use the formula:
(20C0) = 20! / (0! * (20-0)!)

Using this formula, we can calculate (20C0) = 1.

Therefore, P(0) = 1 * (0.05)^0 * (0.95)^20 ≈ 0.358

2. Probability of 1 defective bulb (x = 1):
P(1) = (20C1) * (0.05)^1 * (0.95)^(20-1)

Using the combination formula, we can calculate (20C1) = 20.

Therefore, P(1) = 20 * (0.05)^1 * (0.95)^19 ≈ 0.377

3. Probability of 2 defective bulbs (x = 2):
P(2) = (20C2) * (0.05)^2 * (0.95)^(20-2)

Using the combination formula, we can calculate (20C2) = (20!)/(2! * (20-2)!) = 190.

Therefore, P(2) = 190 * (0.05)^2 * (0.95)^18 ≈ 0.286

To calculate the probability that a random sample of 20 bulbs will contain at most 2 defective bulbs, we add up the individual probabilities:
P(at most 2) = P(0) + P(1) + P(2)

P(at most 2) ≈ 0.358 + 0.377 + 0.286 ≈ 0.821

Therefore, the probability that a random sample of 20 bulbs will contain at most 2 defective bulbs is approximately 0.821.