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

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- Apr 30th 2008, 09:11 AMgarrett88Probability - 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 - Apr 30th 2008, 09:18 AMKai
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 - Apr 30th 2008, 09:24 AMTheEmptySet

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

$\displaystyle \left( \frac{1}{2} \right)^8+8\left( \frac{1}{2} \right)^8=9\left( \frac{1}{2} \right)^8=\frac{9}{256}$ - Apr 30th 2008, 09:30 AMKai
- Apr 30th 2008, 01:11 PMTheEmptySet
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. - Apr 30th 2008, 01:26 PMjanvdl
**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} $ $\displaystyle + {8 \choose 4} \left( \frac{1}{2} \right) ^{4} \left( \frac{1}{2} \right) ^{4} + {8 \choose 5} \left( \frac{1}{2} \right) ^{5} \left( \frac{1}{2} \right) ^{3} + {8 \choose 6} \left( \frac{1}{2} \right) ^{6} \left( \frac{1}{2} \right) ^{2} ]$