# Homework Help: Two more hard probability questions

1. Dec 12, 2004

### ms. confused

Okay, it seems like I've tried everything to solve these two problems, and I still am not getting the answers right. Can someone please help me out?

1. A basketball player makes 80% of her foul shots.
a) What is the probability that she will make 8 of the next 10 foul shots?
b) What is the probability that she will make at least 8 of the next 10 foul shots?

2. Physicians estimate that the likelihood of survival during a particular operation is 65%. What is the probability that 6 or 7 of the next 7 patients survive this operation?

2. Dec 12, 2004

### quasar987

Do you know these are done using the Bernoulli thingy?

Last edited: Dec 12, 2004
3. Dec 12, 2004

### ms. confused

:uhh: Can you be more specific? Is this another law that I don't know about?

4. Dec 12, 2004

### quasar987

for exemple for #2:

R: the person survives. P(R) = 0.65 = p

E: the person dies. P(E) = 0.35 = p - 1

Let X be the number of persons who survive out of 7 persons. then X = {0,1,2,...,7}

I'm not sure from there.. it's been a while since I did prob. But I hope I set you on the right track.

5. Dec 12, 2004

### ms. confused

What's the equation? What part of it is X? Is a binomial probability?

6. Dec 12, 2004

### Pyrrhus

Yes it's a binomial distribution.

$$b(x;n,p) = \left(\begin{array}{c}n&x\end{array} \right) p^{x} (1-p)^{n-x} x=0,1,2, n$$

where n is the total of trys, x the value asked, and p is the probability of success.

Last edited: Dec 12, 2004
7. Dec 12, 2004

### quasar987

The word I was looking for was "Bernoulli trial".

See this page

http://www.mathwords.com/b/bernoulli_trials.htm

The example is exactly similar to what you're looking for, except you'll have to add the probability of having exactly 7 sucess and 6 sucess.

But I'm pretty sure there was a neater way of doing it that involved a sum...

8. Dec 12, 2004

### quasar987

And for #1 b) the "at least" means you'll have to compute the binomial sum 3 times (for 8, 9 and 10) and add the probabilities you find.

Good luck, I'm going to bed.

9. Dec 12, 2004

### ms. confused

Ok, well thank you sooooo much for your input! I see it now! Thanks a lot!

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