Need help with a probability problem that is causing me some difficulty.
A coin is tossed 8 times. What is the probability of more than six tails?
Thanks.
garrett88
This is a binomial probibility
The prob of 8 tails is$\displaystyle \left( \frac{1}{2} \right)^8$
and there is only one way this can happen.
The prob of 7 tails is $\displaystyle \left( \frac{1}{2} \right)^7\left( \frac{1}{2} \right)^1$ but this can happen 8 ways
$\displaystyle \left( \frac{1}{2} \right)^8+8\left( \frac{1}{2} \right)^8=9\left( \frac{1}{2} \right)^8=\frac{9}{256}$
Let H be heads and T be tails If you toss a coint eight times this is the possible ways you can get 1 head and 7 tails
{H,T,T,T,T,T,T,T}
{T,H,T,T,T,T,T,T}
{T,T,H,T,T,T,T,T}
{T,T,T,H,T,T,T,T}
{T,T,T,T,H,T,T,T}
{T,T,T,T,T,H,T,T}
{T,T,T,T,T,T,H,T}
{T,T,T,T,T,T,T,H}
There are a total of 8 ways
You could also use the formula for binomial prob
$\displaystyle P(x) =\binom{8}{x}P^{x}(1-P)^{8-x}$
$\displaystyle P(7)=\binom{8}{7} \left( \frac{1}{2} \right)^7 \left( \frac{1}{2}\right)^1=8 \cdot \left( \frac{1}{2} \right)^8 $
$\displaystyle P(8)=\binom{8}{8} \left( \frac{1}{2} \right)^8 \left( \frac{1}{2}\right)^0=\left( \frac{1}{2} \right)^8 $
This gives the same answer.
I hope this clear it up.
Binomial Distribution
Work out the complement of less(or equal to) than 6.
$\displaystyle = 1 - [ {8 \choose 1} \left( \frac{1}{2} \right) ^{1} \left( \frac{1}{2} \right) ^{7} + {8 \choose 2} \left( \frac{1}{2} \right) ^{2} \left( \frac{1}{2} \right) ^{6} + {8 \choose 3} \left( \frac{1}{2} \right) ^{3} \left( \frac{1}{2} \right) ^{5} $ $\displaystyle + {8 \choose 4} \left( \frac{1}{2} \right) ^{4} \left( \frac{1}{2} \right) ^{4} + {8 \choose 5} \left( \frac{1}{2} \right) ^{5} \left( \frac{1}{2} \right) ^{3} + {8 \choose 6} \left( \frac{1}{2} \right) ^{6} \left( \frac{1}{2} \right) ^{2} ]$