Two more hard probability questions

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Homework Help Overview

The discussion revolves around two probability problems related to binomial distributions. The first problem involves a basketball player's success rate in making foul shots, while the second concerns the survival rate of patients during an operation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the application of Bernoulli trials and binomial distributions to solve the problems. Questions arise regarding the specifics of the equations and the interpretation of the terms involved, such as the meaning of "at least" in the context of the first problem.

Discussion Status

Participants have provided insights into the nature of the problems and the relevant mathematical concepts. Some have shared equations and clarified the need for summation in certain cases, while others express uncertainty about specific steps in the calculations.

Contextual Notes

There is mention of potential gaps in knowledge regarding binomial probability and Bernoulli trials, as well as varying levels of familiarity with the concepts among participants.

ms. confused
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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?
 
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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.

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

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!
 

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