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?

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

i got 10.28 but that rounds up to 10.3

no

explanation:

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