# Coin problem.

• Apr 1st 2009, 03:05 AM
a69356
Coin problem.
Hi,

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.

Thanks,
Ashish
• Apr 1st 2009, 03:51 AM
Plato
Quote:

Originally Posted by a69356
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.

This problem takes a bit of work. There are six cases to calculate then add them up.
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 $\displaystyle {8 \choose 3}$ ways to do that.
So the probability of this case is $\displaystyle {8 \choose 3} \frac{1}{2^{10}}$

Now you have five more cases to work out.