simple chance problem

**emptyglasses** · December 17th 2006, 12:33 PM

I need help with figuring out what method to use to figure out this problem..
Here is the problem.

A fair coin was flipped 10 times; compute the chance that the number of heads is less than 4.

Please help me solve this! Thank you.

**Soroban** · December 17th 2006, 01:02 PM

Hello, emptyglasses!

You need the Binomial Probability formula . . . several times.

A fair coin was flipped 10 times.
Compute the chance that the number of heads is less than 4.

"Less than 4 heads" means: 3 heads or 2 heads or 1 head or no heads.
We must find these separate probabilities . . . and add them.

$P(3H)\:=\:\binom{10}{3}\left(\frac{1}{2}\right)^3\ left(\frac{1}{2}\right)^7 \:=\:\frac{120}{1024}$

$P(2H) \:=\:\binom{10}{2}\left(\frac{1}{2}\right)^2\left( \frac{1}{2}\right)^8 \:=\;\frac{45}{1024}$

$P(1H) \:=\:\binom{10}{1}\left(\frac{1}{2}\right)^1\left( \frac{1}{2}\right)^9 \:=\:\frac{10}{1024}$

$P(0H)\:=\:\binom{10}{0}\left(\frac{1}{2}\right)^0\ left(\frac{1}{2}\right)^{10}\:=\:\frac{1}{1024}$

Therefore: . $P(H < 4) \;=\;\frac{120}{1024} + \frac{45}{1024} + \frac{10}{1024} + \frac{1}{1024} \;=\;\frac{176}{1024} \;=\;\boxed{\frac{11}{64}}$

**emptyglasses** · December 17th 2006, 01:30 PM

Is there a way of figuring this out without doing all of that computing?
Another question asked is A fair coin was flipped 100 times, compute the chance that the number of heads is less than 40. I can't imagine doing an equation for every single one through 40!

**ThePerfectHacker** · December 17th 2006, 01:36 PM

Originally Posted by emptyglasses

Is there a way of figuring this out without doing all of that computing?
Another question asked is A fair coin was flipped 100 times, compute the chance that the number of heads is less than 40. I can't imagine doing an equation for every single one through 40!

I devised my own approach to this problem.
Such that can be programmed into a computer.
I presume there is a way to do it via normal distribution but I never learned probability.

My approah is to find,
$\int_0^{40} \frac{100!}{40!\cdot \Gamma (41-x)}.5^{100} dx$
Where, $s>0$
$\Gamma (s)=\int_0^{\infty} e^{-t}t^{s-1}dt$ (infamous Gamma function).

I know it looks ugly but programs can do this easily. Taking integral by Simpon's rule. Thus, all you need to do is tell your computer to find this integral.

**eureka** · December 17th 2006, 06:53 PM

Originally Posted by emptyglasses

Is there a way of figuring this out without doing all of that computing?
Another question asked is A fair coin was flipped 100 times, compute the chance that the number of heads is less than 40. I can't imagine doing an equation for every single one through 40!

Hi emptyglasses;

Could i suggest to use Normal distribution method. Normal distribution can be used as an approximation to the binomial distribution (as suggested by soroban) when:
- Number of trail is large (greater than 30)
- probability of success at any trail is not too small or large (the closer to 0.5 the better)

Here you go:

First you will need to find the expected mean: number of trail x probability of success at any trail = 100 x 0.5 = 50

Next, find the standard deviation = squaroot 100 x 0.5 x 0.5 = 5

Then you can find the probability by; P(z < (39.5 - 50)/5) = P(z < -2.1) =0.0179 (find from Normal disribution tables)

**CaptainBlack** · December 19th 2006, 04:21 AM

Originally Posted by ThePerfectHacker

I devised my own approach to this problem.
Such that can be programmed into a computer.
I presume there is a way to do it via normal distribution but I never learned probability.

My approah is to find,
$\int_0^{40} \frac{100!}{40!\cdot \Gamma (41-x)}.5^{100} dx$ .

Are you sure that is what you use? What about:

$\int_0^{40} \frac{100!}{\Gamma(x+1)\cdot \Gamma (101-x)}(0.5)^{100} dx$ ?

RonL

**CaptainBlack** · December 19th 2006, 04:32 AM

Originally Posted by CaptainBlack

Are you sure that is what you use? What about:

$\int_0^{40} \frac{100!}{\Gamma(x+1)\cdot \Gamma (101-x)}(0.5)^{100} dx$ ?

RonL

It is a nice excecise to compare the normal approximation, the above
approximation and the exact value. These are 0.0179, 0.0227 and 0.0176
respectivly.

Though if we apply a continuity correction to the integral above so we
integrate from -0.5 to 39.5 the integral changes from 0.0227 to 0.0178
which is indistinguishable from the normal approximation in this case.

RonL

**ThePerfectHacker** · December 19th 2006, 05:54 AM

Originally Posted by CaptainBlank

It is a nice excecise to compare the normal approximation, the above
approximation and the exact value. These are 0.0179, 0.0227 and 0.0176
respectivly.

Though if we apply a continuity correction to the integral above so we
integrate from -0.5 to 39.5 the integral changes from 0.0227 to 0.0178
which is indistinguishable from the normal approximation in this case.

RonL

Thank you for the correction. But my purpose in the post is that it amazingly turns out that we use integration. The reason why the Gamma function appears, as you guessed, is to generalize the factorial. I have a question, is this approach, known? In probability theory.

**CaptainBlack** · December 19th 2006, 06:10 AM

Originally Posted by ThePerfectHacker

Thank you for the correction. But my purpose in the post is that it amazingly turns out that we use integration. The reason why the Gamma function appears, as you guessed, is to generalize the factorial. I have a question, is this approach, known? In probability theory.

I have a sneaking suspicion that it is virtualy identical to the normal approximation (certainly in this case it appears to be)

I think the problem is that if you have the computational resources to
to the integral, you also have the resources to do the exact calculation.

RonL

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