Two more hard probability questions

  • Thread starter Thread starter ms. confused
  • Start date Start date
  • Tags Tags
    Hard Probability
AI Thread Summary
The discussion revolves around solving two probability problems involving binomial distributions. The first problem requires calculating the probability of a basketball player making 8 out of 10 foul shots, and the probability of making at least 8. The second problem involves determining the survival probability of 6 or 7 out of 7 patients during an operation with a 65% survival rate. Participants clarify the use of Bernoulli trials and the binomial probability formula to approach these problems. The conversation concludes with appreciation for the guidance provided.
ms. confused
Messages
90
Reaction score
0
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?
 
Physics news on Phys.org
Do you know these are done using the Bernoulli thingy?
 
Last edited:
quasar987 said:
Do you know these are done using the Bernouilli thingy?

:rolleyes: Can you be more specific? Is this another law that I don't know about?
 
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.
 
What's the equation? What part of it is X? Is a binomial probability?
 
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:
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...
 
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.
 
Ok, well thank you sooooo much for your input! I see it now! Thanks a lot!
 
Back
Top