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May 16th, 2011, 05:03 PM  #1 
Senior Member Joined: Mar 2011 Posts: 105 Thanks: 0  Binomial Theorem (probability)
If Jose Guillen were to maintain his 0.337 average, what is the probability that Jose would get exactly 4 hits in the next 5 at bats? If I solved by myself, I would solve like: (0.337)^4 x 0.663 But my teacher use Binomial Theorem.... IT's too hard to understand.. Can somebody explain why he use Binomal THeorem to solve it, please?? and where can Binommial THeorem be used?? Give me some example about Binomial theorem and Probability, please 
May 16th, 2011, 05:25 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Binomial Theorem (probability)
To solve the question, you want to use: Which is the 5th term from the binomial expansion: The first term represents the probability of getting no hits, the second term 1 hit, etc. and the sum of all these terms must equal 1. For n trials, we have: So, just remember the binomial coefficient in front of each term. 
May 16th, 2011, 06:25 PM  #3  
Senior Member Joined: Mar 2011 Posts: 105 Thanks: 0  Re: Binomial Theorem (probability) Quote:
why the probablity is 5C4 x (0.337)^4(10.337)^1  
May 16th, 2011, 07:56 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Binomial Theorem (probability)
The binomial theorem is closely tied to certain concepts in probability theory, particularly the binomial probability formula and normal distribution. Your stats textbook should give the development of the theory. To answer your question, "why is the probability 5C4 x (0.337)^4(10.337)^1?" we observe that there are 5 ways for José to get exactly 4 hits in the next 5 at bats. Let H represent hit and M represent miss, and the five ways are: way..........probability MHHHH.....(1  0.337)(0.337)(0.337)(0.337)(0.337) = (0.337)^4(10.337)^1 HMHHH.....(0.337)(1  0.337)(0.337)(0.337)(0.337) = (0.337)^4(10.337)^1 HHMHH.....(0.337)(0.337)(1  0.337)(0.337)(0.337) = (0.337)^4(10.337)^1 HHHMH.....(0.337)(0.337)(0.337)(1  0.337)(0.337) = (0.337)^4(10.337)^1 HHHHM.....(0.337)(0.337)(0.337)(0.337)(1  0.337) = (0.337)^4(10.337)^1 You see there are five possible places the one miss can occur. Thus, adding the individual probabilities from each separate scenario, we get: P(X) = nCr(5,4)(0.337)^4(10.337)^1 = 0.042756597377715 The binomial probability formula is used when there are two possible outcomes to a trial, as in this case, either hit or miss, and is given by: where is this case we have: n = 5 x = 4 p = 0.337 Let's look at this in more general terms. Suppose we wish to find out how many subsets of cardinality m can be constructed from an original set of cardinality n, where m ? n. Let's begin by setting up a placeholder for one such subset: {__,__,__,...,__,__,__} Note that there are m blank spaces, corresponding to the cardinality of the subset. If we were to construct a subset at random from the original set, we would have n choices for the first blank space. In the next space, we would have n  1 choices, in the next space n  2 choices and so on down to a number of choices represented by the value n  (m  1). This means that the total number of unique permutations, or sequences of elements, we could encounter would be given by: Using factorial notation, we can write this as: This is the number of permutations that arise when taking n things m at a time. In basic set theory, two sets are equivalent if they contain the same elements, regardless of the order in which they happen to be listed. Suppose that now we have our subset of cardinality m. How many ways can the same group of elements be ordered? For the first space, we would have m choices, for the second m  2, and so on until there is 1 choice for the last space. In factorial notation this is m!. So we take the number of permutations, divide by the number of ways the elements can be arranged, and arrive at the number of combinations: 
May 17th, 2011, 08:57 AM  #5  
Senior Member Joined: Mar 2011 Posts: 105 Thanks: 0  Re: Binomial Theorem (probability) Quote:
I"m so confused because at the beginning of chapter: I have a problem question like " what is probablity to get 4 heads in 5 times toss a coin?" and the random probability for 1 head in 1 times toss is : 1/2 so the probablity to get 4 heads in 5 times toss a coin is : 5 x (1/2)^5 or different method is : (5C4)/(2^5) and I think it's same here......I"m stil confused Mr. MarkFL Please help me  
May 17th, 2011, 10:31 AM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Binomial Theorem (probability)
Using the binomial probability formula, we have: 
May 17th, 2011, 12:38 PM  #7  
Senior Member Joined: Mar 2011 Posts: 105 Thanks: 0  Re: Binomial Theorem (probability) Quote:
oh my godness, I got it. I got it Thank you so much but I think BInomial THeorem is too complicated, right??? My way is easier to understand, right? ( my way is 5C4/(2^5)) Can you give me another example about BInomial THeoram, please???  
May 17th, 2011, 12:49 PM  #8 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Binomial Theorem (probability)
You are using the binomial theorem, but you are just writing (1/2)^5 as 1/(2^5). Using the binomial theorem is the best, most succinct way to do these problems. Here is an article on binomial probability: http://en.wikipedia.org/wiki/Binomial_probability 
May 17th, 2011, 03:53 PM  #9 
Senior Member Joined: Apr 2011 From: USA Posts: 782 Thanks: 1  Re: Binomial Theorem (probability)
I'm gonna take a crack at this one. One problem with a coin toss is that the probability of success and the probability of failure are the same. Any time two numbers are the same, it's easy to get mixed up which .5 is which. That is, it's a coincidence with coins that: happens to also equal . But that wouldn't work for the original problem posted. It's also a coincidence that it's such a nice neat fraction of 1/2. daivinhtran, what you left out of your original solution is the combination in the front of the equation. I'm going to try to avoid all the fancy language. (Which is easy to do since I don't understand it myself. ) Here's my layman's way of looking at it: Let's pretend for the moment that we don't care what order the hits come in. The probability of the first being a hit is .337. The probability of a second is another .337. The probabiliy of the third is .337. Etc. If we're going to have four hits, then (.337)(.337)(.337)(.337). Hence, Then there's one left to be a miss. The probability of a miss is 1  .337, or .663. Since there's only one miss, we only have one .663. So it's Granted, we know that's just .663. But personally, I would get in the habit of putting the exponent on it, since it won't always be a 1, just to make sure you have it in your head what is supposed to happen there. This gives us what you did: But what you haven't accounted for is the fact that those four hits don't have to be the first four times at bat. You have four hits and one miss, out of five times at bat. And how many different ways can that happen? Well, as MarkFL was showing, HHHHM, HHHMH, etc. There's where the combination comes from. We have to do the for every combination that exists. So we have to multiply by the 5C4: 
May 17th, 2011, 04:05 PM  #10 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Binomial Theorem (probability)
Well stated, Erimess! 

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