Binary probability puzzle

[Duratan]

So I was confused if I should post this to math or to brain teasers, if I posted in the wrong spot I apologize, I got this probability question and I couldn't figure out the answer could some help me with the needed formula to solve it. 

The question is: given a binary string that contains 11 digits what is the probability that there is exactly four 1s in the number.  I can solve half of it, I know that 11 digits means 2^11 or 2048 total numbers, but I can't figure out how to find out how many contain exactly four 1s through a formula.

[Strange]

Is this homework? (There is a section of the forum for that, and special rules.)

I think the approach I would take is:

What is the probability of there being exactly one 1? Well, it can be in any position, so it is 1 in 11.

What is the probability of there being exactly two 1s? Well, the first can be in any position, so it is 1 in 11; the second can be in any of the remaining positions so it is 1 in 10. So, total probability is (I think!) 1/11 x 1/10.

And so on ...

[Duratan]

no this is not homework, It's just a puzzle I came across and couldn't figure it out, I dont think that is right, I think your doing the probability that there is a 1 in 11 then one in 10... that doesn't fit this scenario where we want the number of permutations of an eleven digit binary sequence that contains four 1s (the answer  will be that divided by 2048) 

 

[mathematic]

Use binomial:  P_k=(2^-n)ⁿC_k where n is total number of items (11 in your case) and k is the particular number desired (4 in your case).

Edited May 17, 2018 by mathematic
format problem

[J.C.MacSwell]

2 hours ago, Strange said:

Is this homework? (There is a section of the forum for that, and special rules.)

I think the approach I would take is:

What is the probability of there being exactly one 1? Well, it can be in any position, so it is 1 in 11.

What is the probability of there being exactly two 1s? Well, the first can be in any position, so it is 1 in 11; the second can be in any of the remaining positions so it is 1 in 10. So, total probability is (I think!) 1/11 x 1/10.

And so on ...

You might want to rethink that

[Strange]

7 minutes ago, J.C.MacSwell said:

You might want to rethink that

Yes. You are right. As the OP says, I was thinking of permutations, not probabilities...

[J.C.MacSwell]

8 minutes ago, Strange said:

Yes. You are right. As the OP says, I was thinking of permutations, not probabilities...

I thought to myself how long it might take you to realize it if I brought it up...I figured under 10 seconds...so I saved some typing

Edited May 17, 2018 by J.C.MacSwell

[Duratan]

thanks mathematic