Probability problem

[virtualmaster999]

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Hey everyone

Quick question on probability:
"Prob of getting 4 heads with 6 flips of a coin?"
I'm confused on how to go about this. Apparently you are supposed to do 6!/4! And that number gets divided by totally number of possibilities

Is there an easy way to go about this?

Thanks!


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[dawg289]

Total number of possibilities
Imagine you start with HHHHHH
Then HHHHHT, HHHHTH, etc...
all the way to TTTTTT
Since each spot can be H or T, the total number of possibilities is 2^6=64, which is our denominator.

If we want to figure out the probability of getting 4 heads among 6 spots, we need need to figure out how many ways are there to get 4 heads out of 6 flips.
Examples include HHHHTT, HHHTHT, etc.
Note how they always have 4 H's and 2 T's.
If you imagined spelling different "words", using 4 H's and 2 T's, how many ways would there be?
We have 6 letters (so 6 factorial ways of ordering the letters), but 4 of which are indistinguishable from one another and 2 of which are indistinguishable from one another. We must divide by 4! and 2! because all H's are the same and all T's are the same.
So it would be 6! / ((4!*2!)) = 15
15/64 = our answer.

 

[DAT Destroyer]

Hey everyone

Quick question on probability:
"Prob of getting 4 heads with 6 flips of a coin?"
I'm confused on how to go about this. Apparently you are supposed to do 6!/4! And that number gets divided by totally number of possibilities

Is there an easy way to go about this?

Thanks!


Sent from my iPhone using Tapatalk

First you need to find out how many ways you can pick 4 out of 6. This is a combination: 6C4 = 15. Then multiply by the probability of getting 4 heads out of 6 flops of a coin which is (1/2)^6. Multiply 15 by the probability and that's your answer: 15*(1/2)^6

Hope this helps..

 

[virtualmaster999]

First you need to find out how many ways you can pick 4 out of 6. This is a combination: 6C4 = 15. Then multiply by the probability of getting 4 heads out of 6 flops of a coin which is (1/2)^6. Multiply 15 by the probability and that's your answer: 15*(1/2)^6

Hope this helps..

Is it best to do these types of problems with combinations?


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[FeralisExtremum]

Is it best to do these types of problems with combinations?

Doing these problems with the relevant combination/permutation formula (depending on the question) is usually the ideal approach. Once you're dealing with more than ~3 "events", it gets hard to figure them out manually. If the question asked for the probability of getting one head out of three flips, it's easy enough to figure out the outcomes HTT, TTH, THT in your head. But past that it gets time consuming - doing this manually for 4 heads in 6 flips would be a nightmare.

 

[DAT Destroyer]

Is it best to do these types of problems with combinations?


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You can either do combinations or you can think of it as a linear arrangement with repetition as dawg289 explained. From our experience with students, using combinations usually is the easiest way to do these type of problems.

 

[virtualmaster999]

Thank you to everyone!


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