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سرفصل: بخش ریاضی / سرفصل: احتمالات / درس 13

## بخش ریاضی

14 سرفصل | 192 درس

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## Using Counting Techniques

So far, everything we’ve done has concerned use of the formal algebraic probability rules. Now those rules are very useful, they’re used in many probability problems, but there are some probability problems where you can’t use those we have to use some other techniques.

We have to go back to the basic definition of probability. So here’s the basic definition, fundamental definition of probability is the number of successes over the total number of outcomes. Now in some very simple scenarios, you can just make a list of the cases and then you can count out, okay this many successes, this many total outcomes, it must be this over this.

Usually that’s not gonna be enough to get an answer in a probability question. That might be a way to get to the probability of one piece of the question. Often there’s not gonna be enough to answer a whole question, it turns out that many times when we have to use this approach of counting the number of success, and the total number of outcomes. The numbers are actually too large for us to simply explicitly count and we have to use counting techniques.

Now remember all of those videos we had back in the module on counting techniques. The techniques that we used. There’s the fundamental counting principle, so if the first choice happens in K ways, the second choice happens in n ways, the third choice happens in p ways. Then the total number of ways I could choose, I just multiply all those. That’s the fundamental counting principle.

Then there are permutations, so for example, if I’m putting n things in order, and I wanna know how many different orders can I put those in things in. Well, that would be n factorial, combinations as well, we’re not, we don’t care about orders. So, this is everything about choosing. So for example, I have 10 options and how many ways can I distribute three successes, among those 10 options, something like that.

So, if these are unfamiliar ideas, it would be worthwhile to go back and watch the videos in the counting module to make sure you’re clear on the fundamental counting principle and its relationship to permutations and combination. That’s what we’re going to be talking about applying to probability in this video.

So for example, here’s a question. A committee of three will be selected from eight employees, including Alice and Bob. What is the probability that the chosen committee of three includes Alice and not Bob? So this is a classic counting scenario probability question. The whole idea that we have a certain group and were picking people from this group whenever you start doing that, picking people from the group and then applying various restrictions, getting this person, not getting that person, having that person sit next to that person, all that kind of stuff, that is classic counting techniques scenario.

And so we have to use counting techniques. So let’s think about this. So first of all, we can just ask the question how many groups of three can we make from eight employees? So that would be 8 choose 3, and of course the formula for that would be 8 times 7 times 6, over 3 times 2 times 1.

3 times 2 times 1 is 6, so that cancels. And so there are 56 different groups of 3 that I could possibly choose. So that would be the denominator of the fraction. Usually the denominator is much easier to figure out than the numerator. For the numerator, let’s think about this, what would a successful committee look like, a committee that fits the condition we’re saying here.

Well we have to have Alice on the committee, that’s one of the requirements. And then we have these other two requirements. Now, of course the people. There are eight people altogether, let’s just list these out. So pretend those are the eight people. Their names conveniently begin with the first eight letters of the alphabet.

So, A we’ve already included, and we don’t want to include B, so B’s not gonna be included. So we could pick two people from any of these remaining people. That would be two from six. Two from six could go in to these two slots. So six, choose two.

How many different pairs could we pick from six. That would be 6 times 5 over 2. And of course, that equals 15, that’s the numerator. And so, the probability would be 15 over 56, this is the probability that the chosen committee of three includes Alice and not Bob if we are choosing from this group of eight, assuming that all eight employees are equally likely to be chosen.

So, in Summary we will say, in solving a probability question, sometimes, in fact, usually, it’s advantageous to use the algebraic probability and those rules, sometimes it’s advantageously to use counting techniques. So, as the student, when will you know which to use what that’s what we’ll talk about in the next video.

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