Binominal Distributions Worksheet With Answer Key - Helm 2008 Section 372 Page 3
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Example 7
In a box of floppy discs it is known that 95% will work. A sample of three of the
discs is selected at random.
Find the probability that (a) none (b) 1, (c) 2, (d) all 3 of the sample will work.
Solution
Let the event the disc works be W and the event the disc fails be F . The probability that a
disc will work is denoted by P(W ) and the probability that a disc will fail is denoted by P(F ). Then
P(W ) = 0.95 and P(F ) = 1
P(W ) = 1
0.95 = 0.05.
(a) The probability that none of the discs works equals the probability that all 3 discs fail.
This is given by:
P(none work) = P(F F F ) = P(F ) P(F ) P(F )
as the events are independent
3
= 0.05 0.05 0.05 = 0.05
= 0.000125
(b) If only one disc works then you could select the three discs in the following orders
(F F W ) or (F W F ) or (W F F )
hence
P(one works) = P(F F W )+P(F W F )+P(W F F )
= P(F ) P(F ) P(W )+P(F ) P(W ) P(F )+P(W ) P(F ) P(F )
= (0.05 0.05 0.95)+(0.05 0.95 0.05)+(0.95 0.05 0.05)
2
= 3 (0.05)
0.95 = 0.007125
(c) If 2 discs work you could select them in order
(F W W ) or (W F W ) or (W W F )
hence
P(two work) = P(F W W )+P(W F W )+P(W W F )
= P(F ) P(W ) P(W )+P(W ) P(F ) P(W )+P(W ) P(W ) P(F )
= (0.05 0.95 0.95)+(0.95 0.05 0.95)+(0.95 0.95 0.05)
2
= 3 (0.05) (0.95)
= 0.135375
3
(d) The probability that all 3 discs work is given by P(W W W ) = 0.95
= 0.857375.
Notice that since the 4 outcomes we have dealt with are all possible outcomes
of selecting 3 discs, the probabilities should add up to 1. It is an easy check to verify
that they do.
One of the most important assumptions above is that of independence.The probability
of selecting a working disc remains unchanged no matter whether the previous selected
disc worked or not.
19
HELM (2008):
Section 37.2: The Binomial Distribution
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