Two more hard probability questions

In summary, the basketball player has a 0.8 (80%) chance of making a foul shot, and has a 0.65 (65%) chance of making at least one of her next 10 foul shots.
  • #1
ms. confused
91
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?
 
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  • #2
Do you know these are done using the Bernoulli thingy?
 
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  • #3
quasar987 said:
Do you know these are done using the Bernouilli thingy?

:uhh: Can you be more specific? Is this another law that I don't know about?
 
  • #4
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
What's the equation? What part of it is X? Is a binomial probability?
 
  • #6
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.
 
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  • #7
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
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
Ok, well thank you sooooo much for your input! I see it now! Thanks a lot!
 

1. What is the difference between independent and dependent events?

Independent events are events whose outcomes do not affect each other. This means that the probability of one event occurring does not change based on the outcome of the other event. On the other hand, dependent events are events whose outcomes are affected by each other. This means that the probability of one event occurring is dependent on the outcome of the other event.

2. How do you calculate the probability of multiple independent events occurring?

To calculate the probability of multiple independent events occurring, you can use the formula P(A and B) = P(A) x P(B). This means that you multiply the probabilities of each event occurring individually to find the overall probability of both events occurring together.

3. Can the probability of an event be greater than 1?

No, the probability of an event cannot be greater than 1. A probability of 1 means that the event is certain to happen, while a probability of 0 means that the event will not happen. Any probability greater than 1 would mean that the event is more than certain to happen, which is not possible.

4. What is the difference between permutation and combination?

Permutation refers to the arrangement of a set of objects in a specific order, while combination refers to the selection of a subset of objects from a larger set without regard to order. In other words, permutation takes order into account, while combination does not.

5. How do you calculate the probability of an event not occurring?

To calculate the probability of an event not occurring, you can subtract the probability of the event occurring from 1. This can be represented as P(not A) = 1 - P(A). This is based on the fact that the sum of all possible outcomes must equal 1.

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