Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. There are three doors, and a prize is behind one of them. You choose a door. Now the host, who knows where the prize is, opens another door, and there is no prize behind it. Then he asks you: do you want to switch? Most people say, two doors are left, so it is fifty fifty, and it does not matter. Is that right? Hold on to your answer for now. This chapter is the problem bank for the whole unit on statistics and probability, and this puzzle is one of its Part B problems. So let us see how to use this chapter, and then solve two of its problems together. Let's begin!

0:41How to use this chapter

So, let's open this chapter. Part A has sixteen solved problems, at JEE Main level. Okay? Now, the course says: attempt Part A as a thirty minute test, after you finish Chapters one to six. So it checks the whole unit at once. Right? So, a quick question for you. Sixteen problems in thirty minutes. How much time can you give each problem? Thirty over sixteen is about one point nine; the wavy equals sign means about. So just under two minutes a problem, with a clock running. Got it? Then Part B has fourteen longer problems, at JEE Advanced level. With me? And these problems mix counting, conditioning and expectation, so one question can need two or three ideas. Clear? And Part C has twelve concept checks. They test the ideas that come up most often in statement form, so you answer each one with a reason. Okay?

1:39Part C: answer with a reason

Now, two checks from Part C, and we answer each one with a reason. Here is the first one. Two events, A and B, are mutually exclusive, and P of A and P of B, their probabilities, are both greater than zero; the sign that opens towards the left is the greater-than sign. Can these two events be independent? Let us see. Mutually exclusive means the two events cannot happen together. So P of A intersection B is zero; the cap-shaped sign is intersection, both events at once. Right? But for independent events, P of A intersection B must equal P of A times P of B, and that product is more than zero. Zero on one side, a positive number on the other. So the answer is no, they cannot be independent; the slashed equals sign means not equal. Look, this is a classic trap: mutually exclusive and independent are completely different ideas. Got it? Now the second check, from statistics. Every observation is multiplied by minus two. What happens to the standard deviation, S D for short, written sigma, the Greek letter on screen? Look at the deviations. The mean is also multiplied by minus two, so each deviation, x sub i minus x bar, is multiplied by minus two; the small i below is a subscript, and the bar on top means the mean. Squaring removes the sign, so sigma squared, the variance, becomes four times sigma squared; the small two on top means squared, and the arrow means becomes. With me? So sigma becomes two sigma, which means it is doubled. Oh no, careful here, some students write minus two sigma, but a standard deviation is never negative. The sign does not matter. Okay?

3:32Example 1: is A independent of B, given P(A union B) = 1/2?

Now, Example one, from Part A, marked medium. P of A is one third, P of B is one quarter, and P of A union B is one half; the cup-shaped sign is union, one event or the other. Are the two events independent? Remember, about two minutes. Pause the video and try it first. Done? Okay, let's check. Some students add first: one third plus one quarter is seven twelfths, and that is not one half. So they say, not independent. Is that right? No, no, wait, that is the Part C trap again! Comparing P of A plus P of B with P of A union B only tells you whether the events are mutually exclusive, not whether they are independent. For independence we need P of A intersection B, so use the addition rule. P of A union B equals P of A, plus P of B, minus P of A intersection B, okay? So, put in the numbers. What is P of A intersection B? Rearrange it: one third plus one quarter, minus one half. That is seven twelfths minus six twelfths, which gives one twelfth. Clear? Now the product: one third times one quarter is also one twelfth. Right? Both are one twelfth, so P of A intersection B equals P of A times P of B, exactly as independence needs. So yes, the two events are independent. See, one formula, one check, and done. Got it?

5:09Example 2: three doors, switching wins with probability 2/3

Now, remember our door puzzle? Let's finish it. This one is from Part B, marked hard. There are three doors, and the prize is behind one of them. You choose a door. The host, who knows where the prize is, opens another door with no prize behind it, and offers you a switch. Find the probability of winning if you switch. Pause the video and try it first. Done? Okay, let's check. Two closed doors are left, so the answer is one half, right? No, no, careful, one half needs the two closed doors to be equally likely, and they are not. The host knows where the prize is, so the door he opens gives you information. So start from your first choice. Out of three equally likely doors, it is right with probability one third. Okay? So your first choice is wrong with probability one minus one third, which is two thirds. With me? Now let me ask you this. Suppose your first choice is wrong. Of the other two doors, one has the prize and one is empty. Which door can the host open? Yes, exactly! He cannot open your door, and he will not open the prize door. So he must open the only other empty door, and the door left closed has the prize. Right? So switching wins exactly when your first choice was wrong. And if your first choice was right, switching loses. Clear? So the probability of winning by switching is two thirds, not one half. Puzzle solved: switching doubles your chance, from one third to two thirds! Counting and conditioning in one question, that is what Part B is like. Got it?

6:52Quick revision: independent vs mutually exclusive

So, quick revision. Part A is a thirty minute test, Part B is at JEE Advanced level, and Part C tests ideas in statement form. Independent means P of A intersection B equals P of A times P of B, and that is not the same as mutually exclusive. And classical probability needs equally likely outcomes, so always check that first. Now open Part A, set a thirty minute clock, and start solving, then try the practice problems on this page. You can do this. All the best!