A simple fair coin problem

**director** · May 9th 2010, 11:15 AM

A fair coin is flipped 10 times. What is the probability that the heads turn up 8 or more times (8,9,10)?

My approach:

Probability that H turn up 8 times:
$P(8) = {10 \choose 8} (0.5)^{8} (0.5)^{2}$
Probability that H turn up 9 times:
$P(9) = {10 \choose 9} (0.5)^{10}$
Probability that H turn up 10 times:
$P(10) = {10 \choose 10} (0.5)^{10}$

I'm not 100% sure what to do now. Do I simply add the individual probabilities to get the total?

$P(8,9,10) = P(8) + P(9) + P(10)$

Thanks!

**Anonymous1** · May 9th 2010, 12:07 PM

Originally Posted by director

A fair coin is flipped 10 times. What is the probability that the heads turn up 8 or more times (8,9,10)?

My approach:

Probability that H turn up 8 times:
$P(8) = {10 \choose 8} (0.5)^{8} (0.5)^{2}$
Probability that H turn up 9 times:
$P(9) = {10 \choose 9}(0.5)^{10}$
Probability that H turn up 10 times:
$P(10) = {10 \choose 10} (0.5)^{10}$

I'm not 100% sure what to do now. Do I simply add the individual probabilities to get the total?

$\color{red}{\text{Yes.}}$

$P(8,9,10) = P(8) + P(9) + P(10)$

Thanks!

aNon1

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