9. Suppose that in a production of spark plugs the fraction of defective plugs
has been constant at over a long time and that this process is controlled
every half hour by drawing and inspecting two just produced. Find the
probabilities of getting (a) no defectives, (b) 1 defective, (c) 2 defectives.
What is the sum of these probabilities?
Answers
Given : in a production of spark plugs the fraction of defective plugs
has been constant 2 % at over a long time
process is controlled every half hour by drawing and inspecting two just produced.
To Find : probabilities of getting
(a) no defectives, (b) 1 defective, (c) 2 defectives.
sum of these probabilities
Solution:
the fraction of defective plugs has been constant 2 % at over a long time
=> probability of defective p = 0.02
probability of non defective = 1 - 0.02 = 0.98
n = 2
P(x) = ⁿCₓpˣqⁿ⁻ˣ
probabilities of getting no defectives
x = 0
=> P(0) = ²C₀(0.02)⁰(0.98)² = 0.9604
probabilities of getting 1 defective
x = 1
=> P(1) = ²C₁(0.02)¹(0.98)¹= 0.0392
probabilities of getting 2 defectives
x = 2
=> P(2) = ²C₂(0.02)²(0.98)⁰= 0.0004
sum of these probabilities = P(0) + P(1) + P(2) = 1
0.9604 + 0.0392 + 0.0004 = 1
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