>>>>>i just need part D here<<<<<

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1- heart failure is due to either natural occurrences (87%) or outside factors (13%) outside factors are related to induced substances or foreign objects. Natural occurrences are caused by arterial blockage,disease, and infection. Assume that cause of heart failure between individuals are independent .

A)what is the probability that the first patient with heart failure that enters the emergency room has the condition due to outside factors?
Success probability p=0.13
=(1-0.13)^1-1* 0.13=0.13.

B)what is the probability that third patient with heart failure that enters the emergency room is the first one due to outside factors?
(1-0.13)^3-1 *0.13= 0.098

C) what is the mean number of heart failure patients with the condition due to natural causes that enter the emergency room before the first patient with heart failure from outside factors?

M=E(X)=1/p=7.7

D)what is the probability that the third patient who has heart disease die to outside factors is the 8th patient coming in ?

?

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2- your mother is making "Hareesa" with special nuts. She crushed the nuts into small parts and mixed it well with the dough . She used a rectangle shaped tray and cut the sweet into small equal pieces that are 30 gram each . If on average each piece has 6 nuts, answer the following

A- what is the most likely distribution for the number of nuts in sweet pieces?

B-if you chose 3 pieces and saw no nuts. What is the probability that the forth piece will have 5 nuts?

C-what is the probability of finding 13 nuts in a total of 2 pieces of sweets?

D- what is the probability of finding 13 nuts in 50 grams of Hareesa?

 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #2 Mar 20, 2010, 05:56 AM
1 (d)

Here, your sample consists of 8 patients.
Let N be the natural occurrences and
O be the outside factors.

The question is asking what is the probability of having:
OONNNNNO... Or ONONNNNO, or ONNONNNO, etc.

You have different combinations here, which is $\frac{7!}{2!.5!}$

Now that you know that, it's like the binomial distribution for n=8, p = 0.13, q = 0.87.

Can you give it a try now? The answer is: 0.0230

2. For this one, I prefer not answer because I'm not sure.
 galactus Posts: 2,272, Reputation: 1436 Ultra Member #3 Mar 20, 2010, 07:10 AM
For 1d, be careful. It appears to be a negative binomial.

We use a negative binomial when we want the number of the trial in which the kth success occurs.

$C(n-1, k-1)p^{k}(1-p)^{n-k}$

#2d. Try a Poisson distribution.

Let X be the number of nuts in a 50 gram piece. Then, since there are 6 per 30

Grams, we have ${\lambda}={\alpha}t=6(\frac{5}{3})=10$

$\frac{{\lambda}^{x}e^{\lambda}}{x!}$

$\frac{10^{13}e^{-10}}{13!}=.073$
 DyNaFeN Posts: 7, Reputation: 1 New Member #4 Mar 22, 2010, 09:13 AM
Thank you very much..
If you could figure out 2a ,2b,2c would be greatly appreciated
Because I finished the homework except for question 2 ..

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