5.1The Bernoulli and Binomial Random Variables 145
whereas the corresponding probability for a 3-component system is
(
3
2
)
p^2 (1−p)+p^3
Hence, the 5-component system is better if
10 p^3 (1−p)^2 + 5 p^4 (1−p)+p^5 ≥ 3 p^2 (1−p)+p^3
which reduces to
3(p−1)^2 (2p−1)≥ 0
or
p≥
1
2
(b)In general, a system with 2k+1 components will be better than one with 2k− 1
components if (and only if)p≥^12. To prove this, consider a system of 2k+ 1
components and letXdenote the number of the first 2k−1 that function. Then
P 2 k+ 1 (effective)=P{X≥k+ 1 }+P{X=k}(1−(1−p)^2 )+P{X=k− 1 }p^2
which follows since the 2k+1 component system will be effective if either
(1) X≥k+1;
(2) X=kand at least one of the remaining 2 components function; or
(3) X=k−1 and both of the next 2 function.
Because
P 2 k− 1 (effective)=P{X≥k}
=P{X=k}+P{X≥k+ 1 }
we obtain that
P 2 k+ 1 (effective)−P 2 k− 1 (effective)
=P{X=k− 1 }p^2 −(1−p)^2 P{X=k}
=
(
2 k− 1
k− 1
)
pk−^1 (1−p)kp^2 −(1−p)^2
(
2 k− 1
k
)
pk(1−p)k−^1
=
(
2 k− 1
k
)
pk(1−p)k[p−(1−p)] since
(
2 k− 1
k− 1
)
=
(
2 k− 1
k
)
≥ 0 ⇔p≥
1
2
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