Classical Probability and Counting

Transcript

0:00Introduction

Hello everyone! Let us start with a small bet. You throw an ordinary die four times, and your friend says that at least one six will show up. Now, a six comes only once in six throws, on average. Four throws feel like too few, right? So, would you bet against your friend? Hmm, most people would. By the end of this video, we will count it exactly, and the answer may surprise you. Counting questions like this come up often in JEE, so let's begin.

0:32The classical definition

So, first the setting. We do an experiment, and the capital S on screen is the sample space, the set of all its outcomes. Suppose S has n outcomes, all equally likely, and an event, capital A, contains m of them, okay? Then the probability of A, written P of A, is m over n. The number of favourable outcomes, over the total number of outcomes. That is the classical definition. Really simple, right? But look at the condition: equally likely. Two dice are thrown, and we want the probability that the sum is eight. The sum can be anything from two to twelve, which is eleven values, and eight is one of them. So the answer is one over eleven, isn't it? No, no, careful, that is the trap! Those eleven sums are not equally likely, so we cannot use them as our outcomes. The equally likely outcomes are the ordered pairs. The first die has six faces and the second die has six faces, so there are thirty six pairs. With me? Now count the pairs with sum eight. In each bracket, the first number is the first die and the second number is the second die. Two and six, three and five, four and four, five and three, six and two. That makes five pairs, and two and six is different from six and two, okay? So the probability that the sum is eight is five over thirty six, and not one over eleven. The rule to remember is this: count only outcomes that are equally likely. Got it?

2:03Rules that follow from m over n

Now, a few rules follow straight from m over n. Let me ask you this: can P of A ever be bigger than one? No, never, and it can never be negative either. The count m is at least zero and at most n, so m over n lies between zero and one. The sign that looks like less-than with a line under it means less than or equal to. And the whole sample space S contains all n outcomes, so P of S is n over n, which equals one. Clear? Next, the complement. The event A prime, with the small dash, means not A, so it contains the other n minus m outcomes. So P of A prime equals one, minus P of A, okay? Often, counting what you do not want is much easier. And this gives the most useful rule of all. Questions often ask for at least one six, or at least one red ball. So tell me, what is the opposite of at least one? Yes, exactly, none at all! So P of at least one equals one, minus P of none. Keep this ready, because our bet from the start needs exactly this. Right?

3:16Odds in favour

Now, the last idea for today, odds. The odds in favour of A compare P of A with P of A prime. The colon on screen is read as is to, so we say P of A is to P of A prime, okay? The double arrow on screen means that gives. So if the odds in favour are small a is to small b, then P of A equals small a, over small a plus small b, clear? Now a quick check for you. The odds in favour of an event are three is to five. So the probability is three over five, right? Oh no, careful, that is a very common slip! Three is to five means three shares for and five shares against, so eight shares in all. The probability is three, over three plus five, which is three over eight. Three over five compares P of A with P of A prime, and that is not P of A itself. Got it?

4:09Example 1: at least one six in 4 throws, 671/1296

Now, back to our bet from the start. Here it is as a proper question. We throw a die four times. Find the probability of at least one six. Pause the video and try it first. Done? Okay, let's check. Many students say that each throw gives a six with probability one over six, so four throws give one over six, plus one over six, plus one over six, plus one over six, which is two over three. Is that correct? No, no, wait, it is not! Two throws can both show a six, so simply adding counts those outcomes more than once. So let's count properly. Each throw has six outcomes, so four throws give six to the power four, the small four up top. That is one thousand two hundred and ninety six equally likely outcomes. Okay? Now count the outcomes with no six at all. Each throw must show one of the other five faces, so we get five to the power four, which is six hundred and twenty five outcomes. Right? So P of at least one six equals one, minus five over six in brackets, to the power four. That is one, minus six hundred and twenty five, over one thousand two hundred and ninety six. With me? And that gives six hundred and seventy one, over one thousand two hundred and ninety six. The wavy equals sign means approximately, so this is about zero point five one eight, and the greater-than sign says it is more than one half. So your friend wins this bet more often than not. It is close, but it is more than half, and that surprises most people.

5:44Example 2: 3 balls in 3 boxes, exactly one empty, 2/3

Now let's try a harder one. Three balls are placed at random in three boxes. Find the probability that exactly one box is empty. Pause the video and try it first, and think about which outcomes are equally likely. Done? Okay, let's check. First, the equally likely outcomes. Each ball can go into any of the three boxes, so we get three, times three, times three. That is three to the power three, which gives twenty seven outcomes in all. Okay? Now the favourable ones. Exactly one box is empty, so first we choose which box stays empty. There are three ways to do that. Right? Next, all three balls must go into the other two boxes. Each ball has two choices, so two to the power three, which is eight ways, isn't it? Oh no, careful, two of those eight ways are wrong! If all three balls go into one box, then the other box is empty too, and now two boxes are empty, not exactly one. So we subtract those two cases, and eight minus two gives six. That is the trap in this question. Got it? So the favourable outcomes are three times six, which is eighteen. The small cross means times. The probability is eighteen over twenty seven, and that simplifies to two over three. See how the counting did all the work?

7:06Quick revision: equally likely, 1 minus P(none), odds

So, quick revision. First, P of A is m over n, but only when the outcomes are equally likely. Second, for at least one, find one minus P of none. And third, odds of small a is to small b in favour give P of A equals small a, over small a plus small b, okay? Now go and try the practice problems on this page. Count carefully, and you will do great. All the best!