Eventually tossed a coin many times. Part-I

There are $2^{10} = 1024$ possible outcomes from tossing a (fair) coin $10$ times in a row, all of which have the same probability of occurring.

Now there are $\dfrac{10!}{6!*4!} = 210$ ways of arranging $6$ heads and $4$ tails in a row. Thus the desired probability is $\dfrac{210}{1024} = \dfrac{105}{512}$ , and so $a + b = 105 + 512 = \boxed{617}.$

You can use binomial distribution for this problem. The definition of binomial distribution is that the probability of observing exactly r successes in N independent trials is where p = probability of success and q = probability of failure.

In this question, N = 10, p=q=1/2, r=6. Using the above formula, probability comes out to be 105/512. So the solution is 105+512 = 617.

The probability of flipping a coin exactly k times with n trials = (n choose k)*p^k * (1-p)^(n-k)

(10 choose 6) * (.5)^6 * (1-.5)^(10-6) = 105/512

105 + 512 = 617

We can get 6 sixes through 10C6 and total number of outcomes is 2^10 = 1024. So the probability is = 10c6/1024 = 210/1024 = 105/512. So a = 105 and b = 512. a+b - 617

10C6/2^10=105/512