00:01
For this problem, we know that we have that the general formula is that the probability of k consecutive heads, consecutive heads, is equal to the sum from r equals 1 up to n minus k plus 1 of negative 1 to the power of r plus 1 times n minus r.
00:35
Times k choose r plus 1 over p times n minus r k that's n minus r k choose r minus r minus 1 times p to the power of k r times 1 minus p to the power of r where we have n is equal to 10 k is equal to 4 and p is equal to 1 over 2 now, you can compute this numerous different ways.
01:12
I'm just going to plug it into my software here.
01:15
So to begin, i'll note that n minus k plus 1.
01:19
Actually, i'll go through this in order.
01:22
So we're taking the sum of negative 1 to the power of r plus 1 times.
01:28
Now, for those choose expressions, i put in binomial.
01:33
So our n is 10.
01:34
Then we'd have minus r times k, which is 4.
01:38
Choose r plus 1 over p so that would be 2 times binomial n 10 minus r times 4 choose r minus 1 1 .1, all times p 1 over 2 power of r times k which is 4 so r times 1 minus p which would be 1 over 2 to the power of r times k.
02:13
And we're summing that for r from 1 up to n, which is 10 minus k, 4 plus 1...