Permutation

[elizabeth03]

Find the number of possibilities in the following scenarios.

How many ways can a fair coin be flipped 20 times and result in exactly 13 heads and 7 tails?

First off, I am not even sure this is a permutation. I also can not figure out what the variables n and r are.

My teacher covered all of probability in one class, and I missed the class. I have a test tomorrow and am still very confused. Any help would be greatly appreciated, thank you for your time!!

 

[MarkFL]

This is an application of the binomial probability formula. We have a fair coin, so we are to take from that that the probability of heads is equal to the probability of tails, so both are 1/2. We have 20 trials, hence:

[MATH]P(X)={20 \choose 13}\left(\frac{1}{2}\right)^{13}\left(1-\frac{1}{2}\right)^7={20 \choose 13}\left(\frac{1}{2}\right)^{20}=\,?[/MATH]

 

[LCKurtz]

This problem asks how many ways there are to get exactly 13 heads. You have 20 tosses which you could represent by 20 slots in a row:
__ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

You can get the answer by asking simply: How many ways are there to choose the 13 slots for the heads. Or, alternatively, how many ways to choose 7 slots for the tails. It's a counting problem, not a probability problem.

 

[MarkFL]

This problem asks how many ways there are to get exactly 13 heads. You have 20 tosses which you could represent by 20 slots in a row:
__ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

You can get the answer by asking simply: How many ways are there to choose the 13 slots for the heads. Or, alternatively, how many ways to choose 7 slots for the tails. It's a counting problem, not a probability problem.

Quite right...in my haste, I jumped to the conclusion that a probability was being requested. Thank you for catching that and putting the thread in the proper direction.

 

[mathdad]

This is an application of the binomial probability formula. We have a fair coin, so we are to take from that that the probability of heads is equal to the probability of tails, so both are 1/2. We have 20 trials, hence:

[MATH]P(X)={20 \choose 13}\left(\frac{1}{2}\right)^{13}\left(1-\frac{1}{2}\right)^7={20 \choose 13}\left(\frac{1}{2}\right)^{20}=\,?[/MATH]

There is a chapter or two towards the end of Sullivan's college algebra textbook dedicated to probability. I will need your help in terms of probability questions, which are very fuzzy to most people.