A satellite question (prob. satellite is functioning)

[Guest]

A satellite relies on solar cells for its power and will operate so long as at l;east one of the cells is working. Cells fail independently of one another and the probability that any of the cell fails within one year is 0.8(ie an 80% chance of failing)

i) For a satellite with ten cells, show that there is an 89% chance that the satellite is still operating at the end of the year.

ii) Calculate the minimum number of solar cells required for a satellite to function for at least one year. Assume a 95% probability to be sufficient to meet this requirement.

iii) It is decided that the satellite is to have a 90% chance that it is still operating at the end of two years. Assuming that the probability of failure of a cell in its second year is the same as its probability of failure in its first year, calculate the minimum number of cells required to give the satellite a 90% chance of still being operative at the end of two years.

 

[soroban]

Re: A satellite question

Hello, americo74!

A satellite relies on solar cells for its power and will operate if at least one of the cells is working.
Cells fail independently of one another and the probability that any of the cell fails within one year is 0.8

i) For a satellite with ten cells, show that there is an 89% chance that it is still operating at the end of the year.

ii) Calculate the minimum number of solar cells required for a satellite to function for at least one year.
Assume a 95% probability to be sufficient to meet this requirement.

iii) It is decided that the satellite is to have a 90% chance that it is still operating at the end of 2 years.
Assuming that the probability of failure of a cell in its second year is the same as that in its first year,
calculate the minimum number of cells for the satellite to have a 90% chance of operating at the end of 2 years.

We are given: \(\displaystyle \,P(F)\,=\,0.8\)

(i) The probability that all 10 fail is: \(\displaystyle \(10\,F)\:=\0.8)^{10} \:=\:0.107374182\:\approx\:11\%\)

Therefore: \(\displaystyle \,P(\text{at least 1 operating})\:=\:100\%\,-\,11\% \:=\:89\%\)


(ii) With \(\displaystyle n\) cells: \(\displaystyle \(\text{all fail})\:=\0.8)^n\)
We want this probability to be \(\displaystyle 5\%\) at most.

We have: \(\displaystyle \,(0.8)^n\:\leq\:0.05\)

Take logs: \(\displaystyle \,\:\log(0.8)^n\:\leq\:\log(0.05)\;\;\Rightarrow\;\;n
\cdot\log(0.8)\:\leq\:\log(0.05)\)

Therefore: \(\displaystyle \:n\;\geq\;\frac{\log(0.05)}{\log(0.8)}\)*\(\displaystyle \:=\:13.425...\) . . . \(\displaystyle n \,=\,14\) cells.

* We divided by \(\displaystyle \log(0.8)\), a negative quantity
\(\displaystyle \;\;\;\)hence, the inequality is reversed.


(iii) I'm getting strange results with this part.
I'll have to review my reasoning and logic . . .

 

[mcrae]

i) probability of satelite working at end of year = ?
probability one cell failing is 0.8
probability of all 10 failing is 0.8 ^ 10 = 0.11
1 - 0.11 = 0.89 or 89%

the chances of it operatin gat the end of year two are = operating after one year)^2

you need n cells to have a x% chance of operating at the end of one year
x^2 must = .90 or above
sqrt(.90)=.94868

you need n cells to have a .94868% chance of operating at the end of one year
solve, and n=14

so you need 14 cells to have a 95% chance of success first year as well as for 90% chance of success second year

 

[Guest]

thank you Soroban

Did you work out part 3?

 

[stapel]

Did you work out part 3?

You've been given complete worked solutions on the other two parts. What efforts have you made on this part?

Please be specific. Thank you.

Eliz.

 

[mcrae]

i) probability of satelite working at end of year = ?
probability one cell failing is 0.8
probability of all 10 failing is 0.8 ^ 10 = 0.11
1 - 0.11 = 0.89 or 89%

the chances of it operatin gat the end of year two are = operating after one year)^2

you need n cells to have a x% chance of operating at the end of one year
x^2 must = .90 or above
sqrt(.90)=.94868

you need n cells to have a .94868% chance of operating at the end of one year
solve, and n=14

so you need 14 cells to have a 95% chance of success first year as well as for 90% chance of success second year

that bolded part was how i view part 3, i dont understand how you didnt realize it, just because i didnt write iii) :?