It is known that 10% of the calculators shipped from a particular factory are defective. What is the probability that no more than one in a random sample of four calculators is defective?

Answer: 0.9477

Step-by-step explanation:

p = 10% = 0.1, q = 90% = 0.9, n = 4

The question follows a binomial probability distribution since the experiment (defectiveness of random sample of calculator) is performed more than once (4 calculators are defective).

The question is to find the probability that not more than one in a random sample of 4 calculators is defective that's

p (x≤1) = p (x=0) + p (x=1)

The probability mass function of a binomial probability distribution is given below as

P (x=r) = nCr * p^r * q^n-r

At x = 0

p (x=0) = 4C0 * 0.1^0 * 0.9^4-0

p (x=0) = 4C0 * 0.1^0 * 0.9^4

p (x=0) = 1 * 1 * 0.6561

p (x=0) = 0.6561.

At x = 1

p (x=1) = 4C1 * 0.1^1 * 0.9^4-1

p (x=1) = 4C1 * 0.1^1 * 0.9^3

p (x=1) = 4 * 0.1 * 0.729

p (x=1) = 0.2916

p (x≤1) = 0.6561 + 0.2916 = 0.9477