If 8 fair coins are flipped, what are the odds that there will be more heads than tails?
Dear mathman88,
Since flipping coins in this case are independent events, we can use the Binomial distribution to solve the problem. That is if you define a success as getting a head,
$\displaystyle b(x;n,\theta)=~^{n}C_{x}\theta^{x}(1-\theta)^{n-x}$
where, $\displaystyle x-~number~ of ~succeses$
$\displaystyle n-~number~of~trials$
$\displaystyle \theta-~probability~of~a~success$
Probability of getting more heads than tails= $\displaystyle b(5;8,\frac{1}{2})+b(6;8,\frac{1}{2})+b(7;8,\frac{ 1}{2})+b(8;8,\frac{1}{2})$
Hope you can continue.
I like Sudharaka's solution, but here is an alternative approach just to show there is more than one way. Let H be the number of heads, T the number of tails. Then
P(H < T) + P(H = T) + P(H > T) = 1
By symmetry, P(H < T) = P(H > T), so
2 P(H > T) + P(H = T) = 1
It's pretty easy to find P(H = T) because this is just the case H = T = 4. Then use the equation above to solve for P(H > T).