Question

A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures.​ Historically, the failure rate for LED light bulbs that the company manufactures is 18​%. Suppose a random sample of 10 LED light bulbs is selected. Complete parts​ (a) through​ (d) below. a. What is the probability that none of the LED light bulbs are​ defective? The probability that none of the LED light bulbs are defective is . 1374. ​(Type an integer or a decimal. Round to four decimal places as​ needed.) b. What is the probability that exactly one of the LED light bulbs is​ defective? The probability that exactly one of the LED light bulbs is defective is . 3017. ​(Type an integer or a decimal. Round to four decimal places as​ needed.) c. What is the probability that four or fewer of the LED light bulbs are​ defective? The probability that four or fewer of the LED light bulbs are defective is nothing. ​(Type an integer or a decimal. Round to four decimal places as​ needed.) d. What is the probability that five or more of the LED light bulbs are​ defective? The probability that five or more of the LED light bulbs are defective is nothing. ​(Type an integer or a decimal. Round to four decimal places as​ needed.)

Answer 1

a)

Here, n = 10, p = 0.18, (1 - p) = 0.82 and x = 0
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 0)
P(X = 0) = 10C0 * 0.18^0 * 0.82^10
P(X = 0) = 0.1374
0

b)

Here, n = 10, p = 0.18, (1 - p) = 0.82 and x = 1
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 1)
P(X = 1) = 10C1 * 0.18^1 * 0.82^9
P(X = 1) = 0.3017
0

c)

Here, n = 10, p = 0.18, (1 - p) = 0.82 and x = 4
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 4).
P(X <= 4) = (10C0 * 0.18^0 * 0.82^10) + (10C1 * 0.18^1 * 0.82^9) + (10C2 * 0.18^2 * 0.82^8) + (10C3 * 0.18^3 * 0.82^7) + (10C4 * 0.18^4 * 0.82^6)
P(X <= 4) = 0.1374 + 0.3017 + 0.298 + 0.1745 + 0.067
P(X <= 4) = 0.9786

d)

Here, n = 10, p = 0.18, (1 - p) = 0.82 and x = 5
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 5).
P(X >= 5) = (10C5 * 0.18^5 * 0.82^5) + (10C6 * 0.18^6 * 0.82^4) + (10C7 * 0.18^7 * 0.82^3) + (10C8 * 0.18^8 * 0.82^2) + (10C9 * 0.18^9 * 0.82^1) + (10C10 * 0.18^10 * 0.82^0)
P(X >= 5) = 0.0177 + 0.0032 + 0.0004 + 0 + 0 + 0
P(X >= 5) = 0.0213