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
