Binomial Probability FormulaDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/