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Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium

Chapter 18, Problem 48

When rolling a pair of dice, there are two ways to get a point total of 3(1+2;2+1) but only one way to get a point total of 2(1+1). How many ways are there of getting point totals of 4 to 12? What is the most probable point total?

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Hello. Everyone in this video we're being told that there are two boxes containing five balls each. The balls are labeled from number one to number 51 ball is drawn from each box that we have being asked here. How many ways can the drawn balls get a sum of four and to determine the most probable total. So here in this part we're gonna go ahead and let ws of four equal for the number of ways To get a sum of four and then we're gonna go ahead and let tees up Mp. You go to the most probable. Hold on. Alright. So I'm gonna sort of make a chart here for the first column. We're going to have this being the sum of balls drawn. So let's say we have maybe one plus one. So we get one the number one ball from each box. Then we'll have a sum of two of the balls being drawn. So that's what this column will go ahead and contain. Then we'll go ahead and write out the possible where's to get this total? So we have a different kind of number of balls being picked. So let's say we have a sum of two. Then the possible ways can just be one and one, it can be two and zero because the balls are labeled from 1-5. And last column here is just going to be the number of ways we go ahead and get this some. Alright, so we're gonna go ahead and start off the total already some being to how we get a sum of two is only one way and that's just one and one. We can see here the number of ways to get this to some is only one way because we have this one here that's going to be one. And moving on to the sum of three, how we get some of three is if we get the balls of one and two or get two and then one. So we have two different combinations here. So we have two different ways now for the sum of four here we get this by adding one and three. We do the reverse of three plus one or even we do two plus two here we have three different ways to do this. So the number of ways is three Continue on for the summer. We're gonna go ahead and go on and on until we get some of 10. So for the sum of five here how we can get this, if we do one plus four, the opposite four plus one then we'll have two plus three and last 33 plus two. The number of ways that we see here is going to equal to four. We're gonna go ahead and scroll down for more space here to continue on for my some we're going to do six. Next. So how we gonna do this is if we add one plus five or the opposite five plus one, we can also do two plus four opposite of that is four plus two. And then of course we also have the three plus three. You can see here that we have five different ways to do this. So that will be five on this column. Moving on to a sum of seven, we can do three plus four opposite of that is four, then three, then we can also do for seven, some of seven is two plus five or five plus two. We can see here that the number of ways to get this sum of seven is equal to four, next moving on to number eight or the sum of eight. We get this with five plus three for the opposite three plus five. And of course we have four plus four as well. You see here that the number of ways to get the sum of eight with these balls is going to be three. Moving on to nine against girl, a little bit murdered down for more space. Alright, so the possible ways to get number nine is going to be four plus five and of course the reverse of that. So five plus four, can you say that? We see that there's only going to be two ways to do this. And lastly we have a sum of 10, how we get this is only one way, just like the sum of two and not being five plus five. And here is just one way to do this. So you can see here that there are three ways of getting a sum of four. So we look for the sum of four, we see that we have three ways of doing so and the most problem total is going to be six since it can be drawn in five different ways. So we have a sum of six here with five being the highest number of ways to get this number. So the likelihood of getting the sum of six of the two balls is equal to the six. Alright, so let's go ahead and write our final answers then where he said that w sub four is equal to the number of ways you guys some before we found out to be three And then T sub MP which is equal to the most probable total is equal to six. Alright, so this is going to be my final answer for this problem. Thank you all so much for about two.