For the k-out-of-n system described in Problem 3.67, assume that each component independently works with probability 12 . 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.

a). The probabilities PW1∣1- out - of -2=23.b). The probabilities PW1∣2- out - of -3=34.

Q.3.17 - A First Course in Probability

Q.3.17

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) $k = 2, n = 3 .$

Step-by-Step Solution

Verified

Answer

a). The probabilities $P (W_{1} ∣ 1 - out - of - 2) = \frac{2}{3}$ .

b). The probabilities $P (W_{1} ∣ 2 - out - of - 3) = \frac{3}{4}$ .

1Step 1: Given Information (Part a)

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

$P (W_{i}) = P_{i}, i \in {1, 2, \dots n}$ .

2Step 2: Explanation (Part a)

$P (W_{1} ∣ 1 - out - of - 2) = \frac{P (1 - out - of - 2 ∣ W_{1}) P (W_{1})}{P (1 - o u t - o f - 2)}$

Because of what $1 -$ out $-$ of $- 2$ means $P (1 -$ out $-$ of $- 2 ∣ W_{1}) = 1$ , and $P (W_{1}) = P = \frac{1}{2}$ , and boxed formula in box $1$ yields:

${P (1 - out - of - 2) = \sum_{1 \leq l \leq 2} (\begin{array}{l} 2 \\ l \end{array}) {(\frac{1}{2})}^{l} {(1 - \frac{1}{2})}^{2 - l} = (\begin{array}{l} 2 \\ 1 \end{array}) {(\frac{1}{2})}^{2} + {(\begin{array}{l} 2 \\ 2 \end{array})}^{\frac{1}{2}})}^{2} = \frac{3}{4}$

Now these probabilities can be substituted into formula above:

$P (W_{1} ∣ 1 - out - of - 2) = \frac{1 \cdot \frac{1}{2}}{\frac{3}{4}} = \frac{2}{3}$ .

3Step 3: Final Answer (Part a)

The probabilities $P (W_{1} ∣ 1 - out - of - 2) = \frac{2}{3}$ .

4Step 4: Given Information (Part b)

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

$P (W_{i}) = P_{i}, i \in {1, 2, \dots n}$ .

5Step 5: Explanation (Part b)

$P (W_{1} ∣ 2 - out - of - 3) = \frac{P (2 - out - of - 3 ∣ W_{1}) P (W_{1})}{P (2 - out - of - 3)}$

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:

$P (2 - o u t - o f - 3 ∣ W_{1}) = P (1 - o u t - o f - 2) \overset{a)}{=} \frac{3}{4}$

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

$P (2 - out - of - 3) = \sum_{2 \leq l \leq 3} (\begin{array}{l} 3 \\ l \end{array}) {(\frac{1}{2})}^{l} {(1 - \frac{1}{2})}^{3 - l} = (\begin{array}{l} 3 \\ 2 \end{array}) {(\frac{1}{2})}^{3} + (\begin{array}{l} 3 \\ 3 \end{array}) {(\frac{1}{2})}^{3} = \frac{4}{8} = \frac{1}{2}$