I keep getting that
$$P(\text{Circuit works}) = P(\geq\text{1 of subcircuit 1}) +P(\geq\text{1 of subcircuit 2}) +P(\geq\text{1 of subcircuit 3}) $$
$$= (0.9*0.1^2 + 0.9^2*0.1 + 0.9^3)*(0.95^2 + 0.95*0.05)*(0.99) = 0.770$$
But this is the wrong answer
$\endgroup$4 Answers
$\begingroup$You almost have the right idea. (Your work/notation is inconsistent. In the first line you have addition, but in the next you are multiplying.) We are looking for the intersection, not the union, of the events. Further, notice that in the first two subcircuits, you neglected to $\color{blue}{\text{count}}$ the different ways that the devices could fail. So for the first one, it should be
$$\sum_{k = 1}^3\color{blue}{\binom{3}{k}}(0.9)^k(1-0.9)^{3-k}$$ and not $0.9\cdot 0.1^2 + 0.9^2\cdot 0.1 + 0.9^3$. Similarly for the second one. Notice that these two follow the binomial distribution.
I will call $S_i$ the number of devices working in subcircuit $i$. Then doing it your way, we have \begin{align*} P(\text{Circuit works})&= P(S_1\geq 1\cap S_2\geq 1\cap S_3\geq 1)\\ &= P(S_1\geq 1)P(S_2\geq 1)P(S_3\geq 1)\\ &=\left[\sum_{k=1}^{3}\binom{3}{k}(0.9)^k(1-0.9)^{3-k}\right]\left[\sum_{k=1}^2\binom{2}{k}(0.95)^k(1-0.95)^{2-k}\right][0.99]\\ &\approx 0.987 \end{align*} where in the second line I used independence.
Alternatively, we can use the complement,
\begin{align*} P(\text{Circuit works})&= P(S_1\geq 1\cap S_2\geq 1\cap S_3\geq 1)\\ &= P(S_1\geq 1)P(S_2\geq 1)P(S_3\geq 1)\\ &=[1-P(S_1=0)][1-P(S_2 = 0)]P(S_3=1)\\ &=(1-(1-0.9)^3)(1-(1-0.95)^2)(0.99)\\ &\approx 0.987. \end{align*}
$\endgroup$ $\begingroup$Going from left to right, the first part of the circuit (3 in parallel) will work unless all 3 devices fail, with probability $$ p_1=1-(1-.90)^3=.999. $$ the second part of the circuit (2 in parallel) will work unless the 2 devices fail, with probability $$ p_2=1-(1-.95)^2=.9975. $$ Now, you need all three part so work simultaneously, with probability $$ p_1\times p_2\times.99=.986537. $$ Therefore (iv) is the correct answer.
$\endgroup$ $\begingroup$I think you have to use the binomial distribution. $P(\text{A=at least one device functions in the first subcircuit})$ $=1-P(\text{no device function in the first subcircuit})$
So $P(A)=1-\binom 30 0.9^0 0.1^3$
$\endgroup$ $\begingroup$For example, the probability that the central pair of parallel devices fails is equal to $(1-0.95)*(1-0.95)$; so the probability that the central pair works is $1-(1-0.95)*(1-0.95)$, and so on.
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