Thread: Probability that the system function

1. 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?

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

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

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

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

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

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

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

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

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

I hope this one is correct . . .
.

3. Thanks mite....

Originally Posted by Soroban
Hello, zorro!

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

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

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

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

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

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

I hope this one is correct . . .
.

Thanks mite ............Even i got the same answer .... cheers