Coin Toss Probability
Date: 05/26/2007 at 09:16:18 From: David Subject: approximate probability Suppose that you toss a balanced coin 100 times. What is the approximate probability that you observe less than or equal to 40 heads? I'm not sure which formula to use. I know the answer is .0228. I assume from Stand. normal prob. chart at -2.00. Do you use the formula: P(A does not occur) = 1 - P(A)?
Date: 05/27/2007 at 19:56:32 From: Doctor Ricky Subject: Re: approximate probability Hi David, Thanks for writing Dr. Math! The probability of an event occurring when there are only two possible outcomes (such as "heads" or "tails" of a coin flip) is known as a binomial probability. As I'm sure you can notice, we cannot just treat each coin flip as an independent trial when calculating the probability, since the probability of two independent events occurring is normally just the product of the individual probabilities. Therefore, if we flip two coins, that would say the probability of getting one heads and one tails would be (1/2)*(1/2) = 1/4, which is incorrect since it is actually 1/2 (which we can find by constructing a tree diagram). [However, the multiplication method would work fine if we considered all the different permutations of our choices, which will be handled shortly.] Therefore, we cannot just multiply the probabilities together. Instead, calculating the likelihood of a binomial event occurring also requires calculating the Bernoulli coefficient of the event. The Bernoulli coefficient is defined as the combination of n trials with k successes, or (n C k), which is: ( n ) n! ( ) = -------- ( k ) k!(n-k)! We multiply this coefficient by the binomial probability, which is: [(p^k)]*[q^(n-k)], where p = probability of success, q = 1 - p, n = number of trials, k = number of successes. This gives us the general formula for binomial probability: ( n ) B(n,k) = ( ) * p^k * q^(n-k) ( k ) While it may seem somewhat tedious, we would use this formula to find your probability of getting less than or equal to 40 heads in 100 tosses. Since it would only give us the probability of one event happening (such as 34 successes in 100 tosses, if we let k = 34 and n = 100), we would use a sum to add up all the individual probabilities to get the total probability. We would also make use of the fact that p = q = 1/2 in the event of our coin-flipping example. That means we can write our equation as: ( 100 ) C(100,k) = ( ) * (1/2)^100 (since we have the same base, we) ( k ) (add the exponents ) Factoring out the (1/2)^100, since it will be the same for all of our terms, we have: 40 C(100,0<=k<=40) = [(1/2)^100] * Sum (100)C(k) k=0 where (100)C(k) is the combination of 100 things taken k at a time, easily plugged into a graphing calculator using the commands: sum(seq(100 nCr K,K,0,40)) This number is much too large to be of any practical value on its own (it has 29 digits), but multiplied by our binomial probability of (1/2)^100, it brings us to our total probability of 0.0284439668. I hope this helped. If you have any questions, please let me know. - Doctor Ricky, The Math Forum http://mathforum.org/dr.math/
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