Results 1 to 3 of 3

Math Help - Probability that the system function

  1. #1
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Probability that the system function

    Question : A system is made up of three components that operate independently from one another. For the system to function, atleast two components must function. We suppose that the probability of functioning of component no.1 is equal to 0.95 that of component no.2 is 0.9 and that of component no.3 is 0.8.
    i) What is the probability that the system functions?
    ii) Given that the system functions, what is the probability that exactly two components function?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,550
    Thanks
    542
    Hello, zorro!

    A system is made up of three components that operate independently from one another.
    For the system to work, at least two components must work.

    Suppose that: . \begin{array}{ccc}P(A\text{ works}) &=& 0.95 \\ P(B\text{ works}) &=& 0.9 \\ P(C\text{ works}) &=& 0.8 \end{array}


    (a) What is the probability that the system works?

    \begin{array}{ccccccccc}P(A \cap B \cap C) &=& (0.95)(0.9)(0.8) &=& 0.684 \\<br />
P(A \cap B \cap C') &=& (0.95)(0.9)(0.2) &=& 0.171 \\<br />
P(A \cap B' \cap C) &=& (0.85)(0.1)(0.8) &=& 0.076 \\<br />
P(A' \cap B \cap C) &=& (0.05)(0.9)(0.8) &=& 0.036 \\ \hline<br />
&& \text{Total:} && 0.967\end{array}

    P(\text{System works}) \:=\:0.967



    (b) Given that the system works, what is the probability that exactly two components work?

    P(\text{Exactly 2 work }|\text{ System works}) \;=\;\frac{P(\text{Exactly 2 work }\wedge\text{ System works})}{P(\text{System works})}

    Numerator: .Exactly 2 components work:
    . . P(A \cap B \cap C\:\!') + P(A \cap B' \cap C) + P(A' \cap B \cap C) \:=\:0.171 + 0.076 + 0.036 \:=\:0.283

    Denominator, from (a): . P(\text{System works}) \:=\:0.967


    Therefore: . P(\text{Exactly 2 work }|\text{ System works}) \;=\;\frac{0.283}{0.967} \;\approx\;0.293


    I hope this one is correct . . .
    .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Thanks mite....

    Quote Originally Posted by Soroban View Post
    Hello, zorro!


    \begin{array}{ccccccccc}P(A \cap B \cap C) &=& (0.95)(0.9)(0.8) &=& 0.684 \\<br />
P(A \cap B \cap C') &=& (0.95)(0.9)(0.2) &=& 0.171 \\<br />
P(A \cap B' \cap C) &=& (0.85)(0.1)(0.8) &=& 0.076 \\<br />
P(A' \cap B \cap C) &=& (0.05)(0.9)(0.8) &=& 0.036 \\ \hline<br />
&& \text{Total:} && 0.967\end{array}

    P(\text{System works}) \:=\:0.967




    P(\text{Exactly 2 work }|\text{ System works}) \;=\;\frac{P(\text{Exactly 2 work }\wedge\text{ System works})}{P(\text{System works})}

    Numerator: .Exactly 2 components work:
    . . P(A \cap B \cap C\:\!') + P(A \cap B' \cap C) + P(A' \cap B \cap C) \:=\:0.171 + 0.076 + 0.036 \:=\:0.283

    Denominator, from (a): . P(\text{System works}) \:=\:0.967


    Therefore: . P(\text{Exactly 2 work }|\text{ System works}) \;=\;\frac{0.283}{0.967} \;\approx\;0.293


    I hope this one is correct . . .
    .

    Thanks mite ............Even i got the same answer .... cheers
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: January 15th 2011, 02:36 PM
  2. Replies: 0
    Last Post: December 6th 2010, 04:20 PM
  3. Probability System of four independent components
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: September 18th 2010, 09:03 AM
  4. Probability of that system function
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: May 4th 2009, 04:14 PM
  5. Probability - system of components
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 16th 2008, 04:42 PM

Search Tags


/mathhelpforum @mathhelpforum