Question 1121297
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<pre>
The probability that all 4 engines are in order and work properly is {{{p^4}}}.


The probability that 3 engines of 4 are in order and work properly is {{{C[4]^3*p^3*(1-p)}}} = {{{4*p^3(1-p)}}}.


The probability that 2 engines of 4 are in order and work properly is {{{C[4]^2*p^2*(1-p)^2}}} = {{{6*p^2*(1-p)^2}}}


The probability to have one of these cases is the sum


P(p) = {{{p^4}}} + {{{4*p^3*(1-p)}}} + {{{6*p^2*(1-p)^2}}}.


They ask you to find minimal "p" such that  P(p) is still greater than  0.995.


It is easy calculation which can be done with MS Excel, for example.


Below is the table which I got in this way:


      Table


 p       P(p)
---------------------
0.99	1.0000
0.98	1.0000
0.97	0.9999
0.96	0.9998
0.95	0.9995
0.94	0.9992
0.93	0.9987
0.92	0.9981
0.91	0.9973
0.90    0.9963
0.89	0.9951    <<<---===
0.88	0.9937
0.87	0.9921
0.86	0.9902
0.85	0.9880
0.84	0.9856
0.83	0.9829


It shows that the minimal "p" under the question is about p = 0.89.
</pre>

I hope that this solution and this explanation is more clear than that of the other tutor.



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<U>Comment from student</U> : &nbsp;Hello Ikleyn, &nbsp;Thanks for your generous help, &nbsp;really appreciate it. 
I don't want to push it hard to greater lengths, &nbsp;but could you please help me figure out a way or a method 
without actually using the brute force technique. &nbsp;That is to say, &nbsp;finding P(p) without trying every single &nbsp;p - from 0.99 
down to 0.88 &nbsp;Anyways, &nbsp;thanks for you help. &nbsp;You have a good time ahead.
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<U>My response</U> :  &nbsp;&nbsp;Yes, &nbsp;such a method does exist.


It is Calculus. &nbsp;You need to take the derivative  &nbsp;P'(p) and equate it to zero.


The derivative will be a polynomial of the degree 3 without the constant term.


After equating to zero you can cancel &nbsp;"p"&nbsp; in degree one - then you will get a quadratic equation.


Then you can solve it using the quadratic formula.


In this way you will get the required "minimal" &nbsp;p &nbsp;"analytically".