Stat 414
Homework 7 Solution
Fall 2008
1. 2.420 (5 points)
P(A) = .4, P(B) = .5, and P(C) = .1
P(at least one of the five is ineffective  A) = 1 – P(none of the five is ineffective  A) = 1 – (1
– .03)
5
= 1  .97
5
.
P(at least one of the five is ineffective  B) = 1 – P(none of the five is ineffective  B) = 1 – (1 – .
02)
5
= 1  .98
5
.
P(at least one of the five is ineffective  C) = 1 – P(none of the five is ineffective  C) = 1 – (1 – .
05)
5
= 1  .95
5
.
P(at least one of the five is ineffective) = P(at least one of the five is ineffective  A)*P(A) + P(at
least one of the five is ineffective  B)*P(B) + P(at least one of the five is ineffective  C)*P(C) =
(.4)( 1  .97
5
) + (.5)( 1  .98
5
) + (.1)( 1  .95
5
)
P(C  at least one of the five is ineffective) = P(at least one of the five is ineffective  C)*P(C) /
P(at least one of the five is ineffective) = .178
2. Using the information on the front inside cover of the text, do 2.52, a,b,d (5 points)
a. (i) Binomial(5,.7). (ii) μ = 3.5, σ
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 Fall '07
 SENTURK,DAMLA
 Probability theory, Quartile, Prize

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