# Coin probability problem!

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• Mar 18th 2010, 09:46 AM
BabyMilo
Coin probability problem!

A fair coin is flipped 60 times, what is the percent chance of flipping heads 25 or fewer times?

would you solve it this way?

Quote:

binomal
B-(60,0.5)
P(X< or = 25)
so it's 60Cr * 0.5^(r) * 0.5^(60-r)
where r=0,1,2,3,4,5,......23,24,25

then that times 100 to get %.

=12.25%
• Mar 18th 2010, 03:12 PM
Archie Meade
Quote:

Originally Posted by BabyMilo

A fair coin is flipped 60 times, what is the percent chance of flipping heads 25 or fewer times?

would you solve it this way?

Yes,

that would be right.

You could also use the Normal approximation to the Binomial.

$\mu=np=30$

$\sigma=\sqrt{npq}=\sqrt{15}$

$Z=\frac{25-30}{\sqrt{15}}=-1.29$

Applying the correction factor $\frac{0.7655}{\sqrt{np}}=0.14$

$Z=-1.29+0.14=-1.15$

which corresponds to 12.5%
• Mar 18th 2010, 03:51 PM
BabyMilo
Quote:

Originally Posted by Archie Meade
Yes,

that would be right.

You could also use the Normal approximation to the Binomial.

$\mu=np=30$

$\sigma=\sqrt{npq}=\sqrt{15}$

$Z=\frac{25-30}{\sqrt{15}}=-1.29$

Applying the correction factor $\frac{0.7655}{\sqrt{np}}=0.14$

$Z=-1.29+0.14=-1.15$

which corresponds to 12.5%

no idea what you did there.
but thanks for confirming.
• Mar 18th 2010, 04:18 PM
Archie Meade
That's ok!

It's a shortcut taken by comparing the Binomial distribution with the Normal distribution.

The calculation can then be done more quickly (especially when we'd have to sum a substantial number of terms of the Binomial expansion).

However, you'd need to have covered the Normal distribution for this.