Binominal Distributions Worksheet With Answer Key - Helm 2008 Section 372 Page 8
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20Example 10
In a box of switches it is known 10% of the switches are faulty. A technician is
wiring 30 circuits, each of which needs one switch. What is the probability that
(a) all 30 work, (b) at most 2 of the circuits do not work?
Solution
The answers involve binomial distributions because there are only two states for each circuit - it
either works or it doesn’t work.
A trial is the operation of testing each circuit.
A success is that it works. We are given P(success) = p = 0.9
Also we have the number of trials n = 30
Applying the binomial distribution P(X = r) =
C p (1
p)
.
30
30
0
(a) Probability that all 30 work is P(X = 30) =
C
(0.9)
(0.1)
= 0.04239
30
(b) The statement that “at most 2 circuits do not work” implies that 28, 29 or 30 work.
That is X
28
P(X
28) = P(X = 28) + P(X = 29) + P(X = 30)
30
30
0
P(X = 30) =
C
(0.9)
(0.1)
= 0.04239
30
30
29
1
P(X = 29) =
C
(0.9)
(0.1)
= 0.14130
29
30
28
2
P(X = 28) =
C
(0.9)
(0.1)
= 0.22766
28
Hence P(X
28) = 0.41135
24
HELM (2008):
Workbook 37: Discrete Probability Distributions
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