Question 346629
For {{{0}}} failures,{{{P=1*(0.93)^10*(0.07)^0=0.48398}}}
For {{{1}}} failure, {{{P=10*(0.93)^9*(0.07)^1=0.36429}}}
For {{{2}}} failures,{{{P=45*(0.93)^8*(0.07)^2=0.12339}}}
Those are the only values we need to solve the problem.
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Now let's work through the problem.
a) Exactly 1 bolt failure, {{{highlight(P=0.36429)}}}
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b) At least {{{2}}} bolt failures, means {{{2}}} failures or more.
In other words, not {{{0}}} or {{{1}}} failures({{{1-(P(0)+P(1))}}}).
{{{P(0)+P(1)=0.48398+0.36429=0.84825}}}
You know that
{{{P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)=1}}}
So then,
{{{P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)=1-P(0)-P(1)}}}
{{{P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)=1-0.84825}}}
{{{P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)=highlight(0.15175)}}}
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c)At most {{{1}}} bolt, means {{{0}}} or {{{1}}} failures.
{{{P(0)+P(1)=highlight(0.84825)}}}
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d)At least {{{1}}} bolt failure, means not {{{0}}} failures ({{{1-P(0)}}}).
{{{P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)=1-P(0)}}}
{{{P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)=1-0.48398}}}
{{{P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)=highlight(0.51602)}}}