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Answer:
Probability of obtaining no more than two defective tubes = 0.816
Step-by-step explanation:
The Probability of obtaining no more than two defective tubes in a randomly selected sample of 15 tubes is obtained using the binomial distribution formula: nCr × p^r × q^(n -r).
Where n is number of samples;
r is maximum number of defective tubes, r ≤ 2;
p is probability of defective tubes = 10% or 0.1
q is probability of non-defective tubes, q = 1 - p
Further explanations and calculations are given in the attachment below:
