our string of m heads can be in n+1 different positions relative to the other n coin tosses. The outcomes of those n tosses do not matter since your question asks for at least a string of m heads. So the probability should be: (n+1)2n/2n+m

So our string of m heads can be in n+1 different positions relative to the other n coin tosses. The outcomes of those n tosses do not matter since your question asks for at least a string of m heads. So the probability should be: (n+1)2n/2n+m.

## Answers ( )

our string of m heads can be in n+1 different positions relative to the other n coin tosses. The outcomes of those n tosses do not matter since your question asks for at least a string of m heads. So the probability should be: (n+1)2n/2n+m

Answer:Step-by-step explanation:So our string of m heads can be in n+1 different positions relative to the other n coin tosses. The outcomes of those n tosses do not matter since your question asks for at least a string of m heads. So the probability should be: (n+1)2n/2n+m.