I stuck in the below question so required assistance :-
Q - A fair coin is tossed 10 times. Find the probability that 2 heads do not occur consecutively.
a) 1/(2^4) b) 1/(2^3) c) 1/(2^5) d) None of these.
Any help would be greatly appreciated.
This problem takes a bit of work. There are six cases to calculate then add them up.
Originally Posted by a69356
You can have anywhere from zero to five heads. You cannot have six. Why.
In the case of having three heads and seven tails, the tails separate the heads.
So there are eight places to place the three heads: _T_T_T_T_T_T_T_.
That means that there are ways to do that.
So the probability of this case is
Now you have five more cases to work out.