Find prob. <= 2 patients not taken inside target time

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sheppady

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Jul 31, 2007
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I'm having some trouble with the following. I'm not sure if I am overcomplicating it:

The probability that a patient complaint is not dealt with in the target time is 0.08. If 36 complaints are received in 2000 - What is the probability that no more than 2 will not be dealt with in the target time?

I have gotten to this:

p(<=2) = p(0)+p(1)+p(2)

p=0.92, q=0.08

p(2 complaints) = 36.(0.92)34 (0.08)2
p(1 complaint) = 36.(0.92)35 (0.08)
p(0 complaints) = (0.92)36

Should I be using the 36 ?

:roll:
 
\(\displaystyle \L \sum\limits_{k = 0}^2 {{ 36 \choose k} \left( {0.08} \right)^k \left( {0.92} \right)^{36 - k} }\)
 
Thank you.

So although I haven't written it out the same as you - my answer is correct (albeit longwinded)?

Deb
 
sheppady said:
The probability that a patient complaint is not dealt with in the target time is 0.08. If 36 complaints are received in 2000 - What is the probability that no more than 2 will not be dealt with in the target time?
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WHAT :!: :?:
Only 3 complaints per month, and these can't be handled within target time?
I thought this could only happen in Ontario :!:
 
I thought this could only happen in Ontario :!:
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No. Certainly not. No one...and I mean, no one.....can beat the US at bureaucracy. Washington!. That's synonymous with inept, bureaucratic, corrupt, and many other adjectives that fit.

Many in the US would like to see universal health care. Sounds like a great idea. But the bureaucratic, corrupt, monster that would create is beyond imagination. Unfortuantely, that sort of thing is impossible here.
 
Just to let you know this original 'probability' question comes from the UK.
 
sheppady said:
Just to let you know this original 'probability' question comes from the UK.
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I realize that. Just making a point. When Denis stated, "only in Ontario", it made me think; No matter how Ontario may be, chances are the US is worse, regarding inept bureacracies. I see it kind of drifted off topic, but I had to. :D
 
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