Probability of broken down cars

Hi everyone,

If i had a fleet of 27 cars and the probability of any one car not working at any one time was 3% (based on the global fleet reliability data from the manufacturer, so there's no relationship between each car in my fleet), what would be the chance of:
All 27 working
26/27 working
25/27 working
....
1/27 working
0/27 working?

Thanks!

Steve

steve.smith5 said:

Hi everyone,

If i had a fleet of 27 cars and the probability of any one car not working at any one time was 3% (based on the global fleet reliability data from the manufacturer, so there's no relationship between each car in my fleet), what would be the chance of:
All 27 working
26/27 working
25/27 working
....
1/27 working
0/27 working?

Thanks!

Steve

Click to expand...

So let me understand this. You want someone to do 28 problems for you and you did not show us that you even tried? Good luck.

Jomo said:

So let me understand this. You want someone to do 28 problems for you and you did not show us that you even tried? Good luck.

Click to expand...

I definitely tried! I think that
Pr[27 working] = 0.97^27 = 43.94%, and
Pr[0 working] = 0.03^27 = 00.00%

Is that right?

But I'm not sure about the ones in the middle. I think it's a combination of nCr or nPr based on my digging around on the internet but I'm not sure.

Perhaps Jomo, you could just tell me the equation and I'll do the work myself !!

Cheers

Steve

These are binomial distribution problems.

Let p = probability of a car not breaking down.

Let q = probability of a car breaking down (so q=1-p)

P( n cars not breaking down)= (27Cn)(p)^n*q^(27-n)

Just use this equation for all your questions by just changing the value for n.