Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. There is a box of ten items, and four of them are defective. You pick three items at random. What is the chance that at least one is defective? Each item is defective with chance four over ten, and you pick three, so three times four over ten gives one point two. So the probability is one point two? Something is badly wrong, right? This chapter has mixed problems in the JEE Main style, on all of statistics and probability. So let us see what it holds, and solve two of its problems together. Let's begin!

0:38What this chapter holds

So, why does this chapter matter? Statistics and probability together often give two or three questions in a JEE Main shift, okay? You have studied five chapters of this unit. So, which kinds of question would you expect in the paper? Let us see how many you guessed. First, a statistics question: a mean and variance reconstruction, or a correction. Reconstruction means rebuilding missing values from the given mean and variance, and correction means fixing a wrongly recorded value. These are in Set one. Right? Second, a counting-based probability, where the real work is counting the outcomes. These are in Set two. And third, often a Bayes or a binomial question. Bayes is in Set two, and the binomial distribution, with random variables, in Set three. Got it? The problems in these sets are original, in the JEE Main style. Each should take just one to two minutes. With me? And the real past questions, from twenty twenty one to twenty twenty six, are at the end of Chapters one to five. Okay?

1:49Tools for statistics and counting

Now, one tool for each type, all from Chapters one to five. The big capital sigma means the sum of, x bar, the x with a bar on top, is the mean, and the small Greek letter sigma is the standard deviation. So, the sum of x equals n times x bar, and the sum of x squared equals n times, in brackets, sigma squared plus x bar squared; the small two on top means squared. So from n, the mean and sigma alone, we rebuild both totals. Got it? After any change in the data, we recompute. The mean, x bar, is the sum of x over n, and the variance, sigma squared, is the sum of x squared over n, minus x bar squared. We divide by n, as our course does. Right? Next, counting. P of A, the probability of A, equals m over n, where m of the n outcomes are favourable, okay? But only when all n outcomes are equally likely. Clear? Now, at least one could mean exactly one, or two, or three, so many cases. Is there a shortcut? Yes, exactly, the complement! P of at least one equals one minus P of none. Count the single case you do not want, and subtract its probability from one. Got it?

3:05Tools for Bayes and the binomial

Now, the tool for Bayes' theorem. The events E one, E two, up to E n, must form a partition: no two overlap, together they cover the whole sample space, and each has a positive probability. The small numbers written low are subscripts, and the three dots mean and so on. Also, P of A must be greater than zero; that is the greater-than sign just before the zero. With me? Then P of E k given A, where the vertical bar is read as given, equals P of E k times P of A given E k, over the sum, over every i, of P of E i times P of A given E i, one product for each branch. In words, the chance that A came through branch k is that branch's share of P of A, okay? And the binomial distribution, B of n comma p. It needs n independent trials, each with two outcomes, success or failure, and the same success probability p every time. Right? Then the mean is n p, and the variance is n p q, where q equals one minus p, the chance of failure. Clear?

4:08Example 1: at least one defective, and why 1.2 is wrong

Now, Example one, from Set two, and it is our puzzle! Ten items in a box, and four of them are defective. Three are chosen at random. Find the probability that at least one is defective. Pause the video and try it first. Done? Okay, let's check. First, our puzzle answer, three times four over ten, where the cross sign means times. Where did it go wrong? Three times four over ten is one point two, bigger than one. But no probability can be bigger than one, so this is wrong. Adding the three chances counts the choices with two or three defective items more than once. Got it? So, let us count properly. The number of ways to choose three items from ten is ten C three, written as ten above three inside round brackets. That is one hundred and twenty, and since the choice is random, all of them are equally likely. Okay? Now use the shortcut. How many choices have no defective item at all? Yes, exactly, all three must come from the six good ones, and six C three is twenty. Right? So P of at least one is one minus twenty over one hundred and twenty. That is one minus one over six, which gives five over six. Our puzzle is solved: five sixths, a proper probability, less than one. Clear?

5:38Example 2: correct mean and variance after a wrong value

Now a harder one, from Set one, marked hard. Twenty five observations have mean thirty, and S D, the standard deviation, is four. One value, forty two, was recorded as twenty four. Find the correct mean and variance. Pause the video and try it first. Done? Okay, let's check. First, rebuild both totals. The sum of x is twenty five times thirty, which is seven hundred and fifty. The variance is four squared, which is sixteen, and thirty squared is nine hundred. So the sum of x squared is twenty five times nine hundred and sixteen, which gives twenty two thousand nine hundred. With me? Read carefully: forty two was recorded as twenty four. So which value sits in our totals now, and must come out? Twenty four, yes, exactly! It went into the calculation, so it comes out, and forty two goes in. Seven hundred and fifty, minus twenty four, plus forty two, equals seven hundred and sixty eight. Okay? Now the squares. Twenty four squared is five hundred and seventy six, and forty two squared is one thousand seven hundred and sixty four. So twenty two thousand nine hundred, minus five hundred and seventy six, plus one thousand seven hundred and sixty four, equals twenty four thousand and eighty eight. Right? The correct mean is seven hundred and sixty eight over twenty five, which gives thirty point seven two. Clear? Now, only one value out of twenty five changed. So is the variance still sixteen? No, it is not, and that is the trap. Recompute with the new totals and the new mean. Twenty four thousand and eighty eight over twenty five is nine hundred and sixty three point five two. Thirty point seven two squared is nine hundred and forty three point seven one eight four. Subtract, and the correct variance is nineteen point eight zero one six. Got it?

7:37Quick revision: three types, rebuild the sums, 1 - P(none)

So, quick revision. This chapter has three types: a mean and variance fix, a counting probability, and a Bayes or binomial question. For statistics, rebuild the sum of x and the sum of x squared, then recompute with the new mean. For at least one, use one minus P of none, and remember, a probability is never above one. Now open Set one, the next page, and start solving, one to two minutes a problem. You can do this. All the best!