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A system has two independent components the first component has eight sub-components which function with probability 0.95 the second has four sub-components which function with probability 0.90.

Six of the eight sub-components of component one must function for component one to function and three of the four sub-components of component two must function for component two to function.

Both components must function for the system to function. What is the probability that the system functions?

The answer is 0.942.

Originally I was thinking to solve this problem I should do the following: A:= the system functions.

$$ p(A)={8 \choose 6 }*(0.95)^6*(0.05)^{2}*{4 \choose 3 }*(0.90)^3*(0.10)^{1} $$

However, this seems to yeild the wrong answer... any ideas?

Thank you!

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There is a minor misinterpretation: it should be at least $6$ of the $8$ sub-components of component one must function, and at least $3$ of the $4$ sub-components of component two must function.

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  • $\begingroup$ Ohh I see. How would I deal with the "at least" case... I can't seem to find an example in my book which goes over a similar example. Thank you for the quick reply! $\endgroup$ – Rob Sep 21 '18 at 2:28
  • $\begingroup$ @Rob You seem to know how to do the computation when it's "exactly $6$." Well "at least $6$" is the same as "exactly $6$ or exactly $7$ or exactly $8$" etc. $\endgroup$ – angryavian Sep 21 '18 at 5:14
  • $\begingroup$ Got it! Thank you so much for the help! $\endgroup$ – Rob Sep 21 '18 at 5:37

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