A coin is tossed M+N times,M>N,then the chance of getting at least consecutive head is
To find the chance of getting at least one consecutive head when a coin is tossed M+N times, where M > N, we can use the principle of inclusion-exclusion and consider the complementary event (not getting any consecutive heads).
Let's start by finding the total number of possible outcomes when a coin is tossed M+N times. Since each toss can result in two possible outcomes (heads or tails), the total number of outcomes is 2^(M+N).
Now, let's calculate the number of outcomes where there are no consecutive heads. We can solve this by considering the possible arrangements of N heads among M tails.
First, we can place N heads in the M possible slots in (M+1)C(N) ways (using combinatorics formula). This accounts for all the possible arrangements of N heads among M tail positions, with at least one tail between each pair of heads.
Next, for each arrangement, there are (M-N+1) positions remaining where we can place the remaining M-N tails.
Therefore, the total number of outcomes without consecutive heads is given by (M+1)C(N) * (M-N+1).
Now, to find the chance of getting at least one consecutive head, we can subtract the probability of not getting any consecutive heads from 1 (as these are complementary events).
The probability is given by: 1 - [ (M+1)C(N) * (M-N+1) / 2^(M+N) ]