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A company is deciding on installing a 3-component or 5-component satellite system in XYZ town. It is known that the system functions on any given day if a majority of the satellite components function on that day. On a rainy day, each of the components independently functions with a probability of 0.4, whereas on a dry day, each independently functions with probability of 0.8. If it is twice as likely to have a rainy day as it is to have a dry day in XYZ town, should the company install the 3-component or 5-component system? Explain.

I would pick the 3 component system because it is twice as likely to rain then be dry. With the probability of the component working on a rainy day being 0.4 wouldn't it make sense to have the 3 component system because then you only need 2/3 functioning instead of 3/5? I'm not sure how to do this mathematically, can someone help me?

3 Answers 3

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3 component system.

2 failures, dry day $= {3\choose2} (0.2)^2 (0.8)$ 3 faiures, dry day $= (0.2)^3$

2 failures, rainy day $= {3\choose2} (0.6)^2 (0.4)$ 3 faiures, rainy day $= (0.6)^3$

chance of failure $= \frac 13 (3\cdot 0.2^2 \cdot 0.8 + 0.2^3) + \frac 23 ( 3 \cdot 0.6^2\cdot 0.4 + 0.6^3)$

5 component system

3 failures, dry day = ${5\choose 3} 0.2^3\cdot 0.8^2$

4 failures, dry day = ${5\choose 4} 0.2^4\cdot 0.8^4$

etc.

Work out probablity of the other events. Then which has the greater chance of failure

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Take the 5-component case: let $k$ be the number of functioning components on any given day. First, apply law of total probability by conditioning on dry/rainy: $$ \mathbb P(k\geq3) = \mathbb P(k\geq3 \mid \text{dry})\mathbb P(\text{dry}) + \mathbb P(k\geq3 \mid \text{rainy})\mathbb P(\text{rainy}) $$ Now take for example $\mathbb P(k\geq3 \mid \text{dry})$. This is $$ \mathbb P(k\geq3 \mid \text{dry}) = \mathbb P(k = 3 \mid \text{dry}) + \mathbb P(k = 4 \mid \text{dry}) + \mathbb P(k = 5 \mid \text{dry}) $$ where $$ k \mid \text{dry} \sim \text{Binom}(5, 0.8) $$ Thus you can calculate the probability of a functioning system.

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The probability of the 3 component system working on a rainy day is the chance of 2 components working or all 3 components working, so that is ${3\choose2}*(0.4)^2*(0.6)+(0.4)^3 = 0.288+0.064=0.352$.

For the 5 components system to work on a rainy day, you need 3,4, or 5 components working, so you get ${{5}\choose{3}}*(0.4)^3*(0.6)^2+ {5\choose4}*(0.4)^4*(0.6)+(0.4)^5= 0.2304+0.0768+0.01024=0.31744$

Similarly, on dry days the probability the 3 component system works is ${3\choose2}*(0.8)^2*(0.2)+(0.8)^3=0.384+0.512=0.896$

And for the 5 component system to work on a dry day the probability is ${5\choose3}*(0.8)^3*(0.2)^2+{5\choose4}*(0.8)^4*(0.2)+(0.8)^5=0.2048+0.4096+0.32768=0.94208$

Since it is twice as likely to rain than not, the chance of rain is $\frac{2}{3}$, and the chance for it to be dry is $\frac{1}{3}$.

So, for the 3 component system to work, the chance is $\frac{2}{3}*0.352+\frac{1}{3}*0.896=\frac{16}{30}$

For the 5 component system it is $\frac{2}{3}*0.31744+\frac{1}{3}*0.94208=\frac{15.7686}{30}$

So, the 3 component system will work more likely than the 5 component system.