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 probabilities.  The third link below is similar to your 
problem, except it uses dice instead of basketballs:

  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/