Question

For the $k$-out-of-$n$ system described in Problem 3.67, assume that each component independently works with probability $\frac{1}{2}$ . Find the conditional probability that component 1 is working, given that the system works, when

(a) $k=1,n=2$;

(b)$k=2,n=3.$

Short answer

a). The probabilities $P\left(W_1\mid 1-\text{out}-\text{of}-2\right)=\frac{2}{3}$.

b). The probabilities$P\left(W_1\mid 2-\text{out}-\text{of}-3\right)=\frac{3}{4}$.

Step-by-step solution

Given Information (Part a)

$W_i$- event that the $i$-th component works, $i\in \{1,2,\dots n\}$.

$P\left(W_i\right)=P_i,i\in \{1,2,\dots n\}$.

Explanation (Part a)

$P\left(W_1\mid 1-\text{out}-\text{of}-2\right)=\frac{P\left(1-\text{out}-\text{of}-2\mid W_1\right)P\left(W_1\right)}{P(1-out-of-2)}$

Because of what$1-$out $-$of $-2$means $P\left(1-\right.$out $-$of $\left.-2\mid W_1\right)=1$, and $P\left(W_1\right)=P=\frac{1}{2}$, and boxed formula in box $1$yields:

${\left.P(1-\text{out}-\text{of}-2)=\sum _{1\le l\le 2}\left(\begin{array}{l}2\\ l\end{array}\right){\left(\frac{1}{2}\right)}^l{\left(1-\frac{1}{2}\right)}^{2-l}=\left(\begin{array}{l}2\\ 1\end{array}\right){\left(\frac{1}{2}\right)}^2+{\left(\begin{array}{l}2\\ 2\end{array}\right)}^{\frac{1}{2}}\right)}^2=\frac{3}{4}$

Now these probabilities can be substituted into formula above:

$P\left(W_1\mid 1-\text{out}-\text{of}-2\right)=\frac{1·\frac{1}{2}}{\frac{3}{4}}=\frac{2}{3}$.

Final Answer (Part a)

The probabilities$P\left(W_1\mid 1-\text{out}-\text{of}-2\right)=\frac{2}{3}$.

Given Information (Part b)

$W_i$- event that the $i$-th component works, $i\in \{1,2,\dots n\}$.

$P\left(W_i\right)=P_i,i\in \{1,2,\dots n\}$.

Explanation (Part b)

$P\left(W_1\mid 2-\text{out}-\text{of}-3\right)=\frac{P\left(2-\text{out}-\text{of}-3\mid W_1\right)P\left(W_1\right)}{P(2-\text{out}-\text{of}-3)}$

If $W_1$occurred, $2$- out $-$of $-3$means that in $W_2,W_3-2$components, 1 more has to be working, and since these are independent, identically distributed:

$P\left(2-out-of-3\mid W_1\right)=P(1-out-of-2)\stackrel{a)}{=}\frac{3}{4}$

And for the denominator use boxed formula in box $1$:

$P(2-\text{out}-\text{of}-3)=\sum _{2\le l\le 3}\left(\begin{array}{l}3\\ l\end{array}\right){\left(\frac{1}{2}\right)}^l{\left(1-\frac{1}{2}\right)}^{3-l}=\left(\begin{array}{l}3\\ 2\end{array}\right){\left(\frac{1}{2}\right)}^3+\left(\begin{array}{l}3\\ 3\end{array}\right){\left(\frac{1}{2}\right)}^3=\frac{4}{8}=\frac{1}{2}$

This is all that is needed to calculate the probability by the stated formula:

$P\left(W_1\mid 2-\text{out}-\text{of}-3\right)=\frac{\frac{3}{4}·\frac{1}{2}}{\frac{1}{2}}=\frac{3}{4}$

Final Answer (Part b)

The probabilities$P\left(W_1\mid 2-\text{out}-\text{of}-3\right)=\frac{3}{4}$.