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### Binomial Probability Formula

```Date: 03/22/2005 at 11:32:33
From: Missy
Subject: probabilities

Sue makes 70% of the free throws she attempts.  She shoots three free
throws in her warmup before a game.  What is the probability that Sue
makes two or more of the three free throws?

I know the answer is .784, but I'm not sure how the book got it.  I
thought you would multiply 2/3 times 3/3, but obviously not.

```

```
Date: 03/23/2005 at 23:15:30
From: Doctor Wilko
Subject: Re: probabilities

Hi Missy,

Thanks for writing to Dr. Math!

Sue makes 70% of the free throws she attempts; therefore 30% of the
time she doesn't make her free throws.

If Sue shoots three free throws, you want to know the probability that
she will make two OR three free throws.  Or in math terms:

P(2 out of 3 shots successful) + P(3 out of 3 shots successful)

Let's sidestep the exact formula for a minute and try to build up the
rationale for how to solve this.

For example, if I wanted to know "What is the probability that Sue
will make 2 free throws in a row followed by a miss?" the answer would
look like this:

P(2 successes followed by 1 miss) = (0.7) * (0.7) * (0.3) = 0.147

This says that Sue will have a 14.7% probability of shooting 2
successful free throws followed by a miss.

This doesn't quite help us (yet) because there are other ways that Sue
could make these shots:

success, miss, success
miss, success, success

There are actually three ways that she could make these shots AND each
one has the same probability.

success, success, miss  = 0.147
success, miss, success  = 0.147
miss, success, success  = 0.147
+ ----------
0.441

So, this finally answers our first part of our question.

P(2 out of 3 shots successful) = 0.441

Now we need to find out P(3 out of 3 shots successful).

Well, if she takes three shots and they are all successful, then the
probability would look as follows:

P(3 out of 3 shots successful) = (0.70) * (0.70) * (0.70) = 0.343

There aren't different arrangements of the shots like before because
all three shots are successful.

So, to answer the main question:

P(2 out of 3 shots successful) + P(3 out of 3 shots successful)

0.441  +  0.343 = 0.784

With Sue's 70% success rate when shooting free throws, she has a 78.4%
probability of making two or more free throws when shooting three free
throws total.

I took some time to develop the rationale of how to solve this
problem.  Once you understand what I did, there is a formula called
the Binomial Probability Formula, which will let you calculate
problems a lot faster than reasoning through it like I did.

The formula looks like:

P(x) = nCx * p^x * q^(n-x)

n = number of trials
x = number of successes among n trials
p = probability of success in any one trial
q = probability of failure in any one trial (q = 1 - p)
nCx = combinations of n items, choose x

If you recognized your problem as a binomial distribution problem (see
links below), then you could go straight to the formula:

P(x) = nCx * p^x * q^(n-x)

P(2 or 3 successful shots) =

P(2 out of 3 shots successful) +  P(3 out of 3 shots successful) =

3C2 * (0.70)^2 * (0.30)^1    +  3C3 * (0.70)^3 * (0.30)^0

0.441  +  0.343 = 0.784

Here's some links from our archives that will elaborate more on

Binomial Experiments
http://mathforum.org/library/drmath/view/63982.html

Binomial Probability
http://mathforum.org/library/drmath/view/56189.html

Probability of Rolling a 2 At Least Twice
http://mathforum.org/library/drmath/view/57596.html

Does this help?  Please write back if you need anything else.  :-)

- Doctor Wilko, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Probability
High School Probability

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