A manufacturer of processors for communication devices has established that the defect probability of its manufacturing process is 0.05. (a) What is the probability of having at the most two defective...

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Let X denote number of defective in n = 30 independent processors, where each
has a probability p = 0.05
We denote this by:
X ∼ Binomial(n = 30, p = 0.05)
A P (X ≤ 2), n = 30, p = 0.05
P (X ≤ 2) = P (0 ≤ X ≤ 2) = P (X = 0, 1, 2)
= P (X = 0) + P (X = 1) + P (X = 2) (because these are disjoint events)
P (X = 0) =
(30
0
)
(0.05)0(1 − 0.05)30 = 1(0.05)0(0.95)30 ≈ 0.214639
P (X = 1) =
(30
1
)
(0.05)1(1 − 0.05)29 = 30(0.05)1(0.95)29 ≈ 0.338903
P (X = 2) =
(30
2
)
(0.05)2(1 − 0.05)28 = 435(0.05)2(0.95)28 ≈ 0.258637
P (X = 0) + P (X = 1) + P (X = 2)
= 0.214639 + 0.338903 + 0.258637 = 0.812179 ≈ 0.812
P (X ≤ 2) = 0.812
Using excel function BinomDist(2, 30,0.05,true) or TI-83/84 function
binomcdf(30,0.05,2), exact answer is 0.8121788131
B P (X > 3), n = 30, p = 0.05
P (X > 3) = 1− P (X ≤ 3)
We first calculate:
P (X ≤ 3) = P (0 ≤ X ≤ 3) = P (X = 0, 1, 2, 3)
= P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) (because these are disjoint
events)
P...