Binary Probability help

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vertex

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Suppose we wanted to create a 12-digit bit string (a bit string is a binary number consisting of only ones and zeros).

How many different 12-digit bit strings are possible?
For this one I did [math]2^{12}[/math] [math] = 4096[/math] Different strings

What is the probability that a12-digit bit string has exactly five ones but begins and ends in zero?(Hint: Remember, all of the different 1’s and 0’s are indistinguishable from one another.)

What is the probability that a12-digit bit string has at least eight ones but begins and ends in zero? (Hint: break it up into cases)

The last 2 have me a bit puzzled as I am not sure where to begin. Could someone tutor me.
 
[QUOTE="vertex, post: 535246, member: 92038"
What is the probability that a12-digit bit string has at least eight ones but begins and ends in zero? (Hint: break it up into cases)[/QUOTE]
Lets play the back-of-the-book game. The textbook has the answers. So we look the answer up.
We see \(\displaystyle\sum\limits_{k = 8}^{10} {\frac{{10!}}{{k!(10 - k)!}}} \), What the heck?
Well you explain it!
 
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Would that be [math]12![/math] / [math]8!(12-8)![/math] = [math]12![/math] / [math]8!4![/math] = [math]12*11*10*9[/math] / [math]4*3*2*1[/math] = [math]11880 / 24[/math] = [math]49.17[/math]
 
vertex said:
What is the probability that a12-digit bit string has exactly five ones but begins and ends in zero?(Hint: Remember, all of the different 1’s and 0’s are indistinguishable from one another.)
Click to expand...
That effectively means you have 10 spaces into which you can put five ones. Work out how many arrangements there are for this and divide by 4096 to get the probability.
(Assuming all 4096 strings are equally likely to occur).
 
Your post appeared after I made my post. Do you know how many combinations there are when choosing 5 objects from 10?
 
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