1(m+n)Cn (1/2)^n (1/2)^m
a coin is tossed (m+n) times where (m>n). find the probability of exactly n consecutive heads.
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8 Answers
62no.. this is not the solution..
This is the probability of getting n heads... not exactly n consecutive heads..
The problem in solving this problem is that there will be a lot of repetition.
Like u could have a block of n heads followed by a tail and another head. so, there will be a lot of recounting.
One way to get across this is to use inclusion exclusion.
There again the issue is that we dont know how far to go...
One could also use the approach of Recurrsion. Other than these I am not sure if any method would work...
I am myself looking forward to the answer if someone could get it correct!
1i know the ans...i can even tel u...if u get me correct method of solving it after looking at ans,.
24it was done by me here
http://targetiit.com/iit-jee-forum/posts/probability-9557.html[6]
66Isn't the question asking for the probability of exactly n consecutive heads, the remaining being tails? In this case, one gets the required probability as
m+12m+n
If the problem asks the probability of n consecutive heads with the others being either head or tail, then also eureka's solution is not entirely valid. Take for example, m=100, n=2. There might be more than one block of 2 heads separated by a sequence of tails. Then the solution given in #6 won't be entirely valid.