If 3% of electric bulb produced by a company are defective find the probability that in a sample 100bulbs
(a) 0 (b)1 (c)2 (d)3 bulbs are defective??
I DON'T KNOW? wHAT IS IT?
a. 0 = 0.0476
b. 1 = 0.1471
c. 2 = 0.2252
d. 3 = 0.2275
use Poisson distribution
To find the probability of a specific number of defective bulbs in a sample, we can use the binomial probability formula:
P(x) = (nCx) * (p^x) * (q^(n-x))
Where:
P(x) is the probability of x defective bulbs
n is the sample size (100 bulbs in this case)
x is the number of defective bulbs
p is the probability of a bulb being defective (3%, which is 0.03)
q is the probability of a bulb not being defective (97%, which is 0.97)
nCx is the binomial coefficient, which represents the number of ways to choose x items out of n
Now, let's calculate the probabilities for each case:
(a) 0 bulbs are defective (x = 0):
P(0) = (100C0) * (0.03^0) * (0.97^(100-0))
= (1) * (1) * (0.97^100)
The value of 0.97 raised to the power of 100 can be calculated using a calculator or a computer program, which gives us the probability.
(b) 1 bulb is defective (x = 1):
P(1) = (100C1) * (0.03^1) * (0.97^(100-1))
= (100) * (0.03) * (0.97^99)
(c) 2 bulbs are defective (x = 2):
P(2) = (100C2) * (0.03^2) * (0.97^(100-2))
= (100 * 99) / (2 * 1) * (0.03^2) * (0.97^98)
(d) 3 bulbs are defective (x = 3):
P(3) = (100C3) * (0.03^3) * (0.97^(100-3))
= (100 * 99 * 98) / (3 * 2 * 1) * (0.03^3) * (0.97^97)
To find the actual values of the probabilities, you will need to substitute the respective values of (100C0), (100C1), (100C2), and (100C3), and perform the calculations using a calculator or a computer program.
Using probability distribution
The probability of gettin