If 3% of electric bulb produced by a company are defective find the probability that in a sample of 100bulbs
(a) 0 (b) 1 (c)2 d(3) bulbs will be defective???
To find the probability for each case, we can use the binomial probability formula. The formula is:
P(X=k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
P(X=k) is the probability of getting exactly k successes,
n is the total number of trials (sample size),
k is the number of successful outcomes (defective bulbs),
p is the probability of success (probability of a bulb being defective),
C(n, k) is the number of ways to choose k successes from n trials.
In this case, n = 100 (sample size) and p = 0.03 (probability of a bulb being defective). Let's calculate the probabilities for each case:
(a) 0 defective bulbs (k = 0):
P(X=0) = C(100, 0) * 0.03^0 * (1-0.03)^(100-0)
C(100, 0) = 1 (since choosing 0 from any number always results in 1)
P(X=0) = 1 * 1 * 0.97^100 ≈ 0.4966
Therefore, the probability of having 0 defective bulbs in a sample of 100 bulbs is approximately 0.4966.
(b) 1 defective bulb (k = 1):
P(X=1) = C(100, 1) * 0.03^1 * (1-0.03)^(100-1)
C(100, 1) = 100 (since there are 100 ways to choose 1 out of 100 bulbs)
P(X=1) = 100 * 0.03 * 0.97^99 ≈ 0.3697
Therefore, the probability of having 1 defective bulb in a sample of 100 bulbs is approximately 0.3697.
(c) 2 defective bulbs (k = 2):
P(X=2) = C(100, 2) * 0.03^2 * (1-0.03)^(100-2)
C(100, 2) = 4950 (since there are 4950 ways to choose 2 out of 100 bulbs)
P(X=2) = 4950 * 0.03^2 * 0.97^98 ≈ 0.1859
Therefore, the probability of having 2 defective bulbs in a sample of 100 bulbs is approximately 0.1859.
(d) 3 defective bulbs (k = 3):
P(X=3) = C(100, 3) * 0.03^3 * (1-0.03)^(100-3)
C(100, 3) = 161700 (since there are 161700 ways to choose 3 out of 100 bulbs)
P(X=3) = 161700 * 0.03^3 * 0.97^97 ≈ 0.0468
Therefore, the probability of having 3 defective bulbs in a sample of 100 bulbs is approximately 0.0468.