# Thread: Probability - coin tosses

1. ## Probability - coin tosses

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

2. Hmm, probability to get a tail is 1/2( am assuming an unbiased coin is being used )

On one throw it is 1/2.

Probability to get more than 6 tails = prob to get 7 tail + prob to get 8 tail

Hopes it helps

3. Originally Posted by garrett88
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 $\left( \frac{1}{2} \right)^8$
and there is only one way this can happen.

The prob of 7 tails is $\left( \frac{1}{2} \right)^7\left( \frac{1}{2} \right)^1$ but this can happen 8 ways

$\left( \frac{1}{2} \right)^8+8\left( \frac{1}{2} \right)^8=9\left( \frac{1}{2} \right)^8=\frac{9}{256}$

4. Originally Posted by TheEmptySet
This is a binomial probibility

The prob of 8 tails is $\left( \frac{1}{2} \right)^8$
and there is only one way this can happen.

The prob of 7 tails is $\left( \frac{1}{2} \right)^7\left( \frac{1}{2} \right)^1$ but this can happen 8 ways

$\left( \frac{1}{2} \right)^8+8\left( \frac{1}{2} \right)^8=9\left( \frac{1}{2} \right)^8=\frac{9}{256}$
I thought the oder of obtaining a head or a tail is not important

So it would be simply (1/2)^7 * (1/2) + (1/2)^8 = 1/128

5. Originally Posted by Kai
I thought the oder of obtaining a head or a tail is not important

So it would be simply (1/2)^7 * (1/2) + (1/2)^8 = 1/128
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

$P(x) =\binom{8}{x}P^{x}(1-P)^{8-x}$

$P(7)=\binom{8}{7} \left( \frac{1}{2} \right)^7 \left( \frac{1}{2}\right)^1=8 \cdot \left( \frac{1}{2} \right)^8$

$P(8)=\binom{8}{8} \left( \frac{1}{2} \right)^8 \left( \frac{1}{2}\right)^0=\left( \frac{1}{2} \right)^8$

This gives the same answer.

I hope this clear it up.

6. Originally Posted by garrett88
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
Binomial Distribution

Work out the complement of less(or equal to) than 6.

$= 1 - [ {8 \choose 1} \left( \frac{1}{2} \right) ^{1} \left( \frac{1}{2} \right) ^{7} + {8 \choose 2} \left( \frac{1}{2} \right) ^{2} \left( \frac{1}{2} \right) ^{6} + {8 \choose 3} \left( \frac{1}{2} \right) ^{3} \left( \frac{1}{2} \right) ^{5}$ $+ {8 \choose 4} \left( \frac{1}{2} \right) ^{4} \left( \frac{1}{2} \right) ^{4} + {8 \choose 5} \left( \frac{1}{2} \right) ^{5} \left( \frac{1}{2} \right) ^{3} + {8 \choose 6} \left( \frac{1}{2} \right) ^{6} \left( \frac{1}{2} \right) ^{2} ]$