Mixed Problems: Set 2 (Probability and Bayes)

Transcript

0:00Introduction

Hello everyone! Let us start with a small mystery. We throw a die. On a six, a ball is drawn from urn A, with two white and three black balls. Otherwise, it comes from urn B, with four white and one black. The ball turns out to be white. Before we saw the ball, the chance of a six was one over six. Now that we have seen white, does that chance go up, or go down? Hold on to your answer; our second example settles it. This is Set two of the mixed problems, on probability and Bayes, in the JEE Main style. So, let's begin.

0:38What Set 2 holds

So, what does Set two hold? Three main groups of problems. First, counting: dice, the letters of a word, people standing in a line. Counting favourable outcomes over all outcomes works only when all the outcomes are equally likely. Okay? Second, independent events. P of A means the probability of A, and the upside down cup is intersection, meaning and. Two events are independent when the chance of both, P of A intersection B, equals P of A times P of B, so we just multiply. And careful, independent is not the same as mutually exclusive. Events that cannot happen together, with non-zero chances, are never independent. Right? And third, conditional probability and Bayes' theorem. Both of today's examples come from here, so let us sharpen these tools first. Clear?

1:31Conditional probability, both ways

Now, conditional probability. P of A given B, where the vertical bar is read as given, equals P of A intersection B, over P of B, okay? It is the chance of A once B has occurred. And the condition on the right: P of B must be greater than zero, because we divide by it. With me? Multiply across, and we get the multiplication theorem. P of A intersection B equals P of B times P of A given B, and also P of A times P of B given A, the other way round. The same overlap, two ways. Okay? Now a question that comes up often. Suppose you know P of A given B, but you need P of B given A, the reverse. How do you get there? Yes, exactly, through the overlap! First find P of A intersection B from the condition you are given, then divide by the other probability. The overlap is the bridge between the two. Got it?

2:31Total probability and Bayes' theorem

Now look at this tree. From the black dot, chance picks one case: E one, E two or E three, with P of E one and so on along the branches; the small numbers are subscripts. Then event A happens, or it does not; here, A prime, with the little dash, means it does not. Okay? The cases E one up to E n, the three dots meaning and so on, form a partition. No two overlap, together they cover the whole sample space, and each has a positive probability. So exactly one of them occurs. Right? Now multiply along each branch that ends in A; those are the products on the right of the tree. Add them all up, and you get P of A, the total. The big capital sigma means add up over i. This is the theorem of total probability. Clear? Now let us turn it around. Suppose we know that event A has happened. How likely is it that it came through case number k? That branch's share of the total, yes! P of E k given A equals P of E k times P of A given E k, over the same sum of all the branches. This is Bayes' theorem: your branch over all the branches ending in A, added up. With me? And two conditions, always. The cases must form a partition, and P of A must be greater than zero, because it sits at the bottom. Okay?

3:54Example 1: find P(B|A) given P(A|B) = 0.4

Alright, Example one, an easy one from the set. P of A is zero point three, P of B is zero point five, and P of A given B is zero point four. Find P of B given A, from these three numbers. Pause the video and try it first. Done? Before we check, one quick question. Many students simply flip it: P of A given B is zero point four, so P of B given A is also zero point four. Is that right? That is not right, and it is a classic trap. We go through the overlap, our bridge. By the multiplication theorem, P of A intersection B equals P of B times P of A given B, and P of B is zero point five, not zero. Zero point five times zero point four, where the cross means times, gives zero point two. Got it? Now divide by P of A, which is zero point three, again not zero. Zero point two over zero point three gives two over three, and that is our answer, P of B given A, right? And look, two over three is not zero point four; the equals sign with a slash through it means not equal. So the order inside the bracket really matters.

5:09Example 2: die and two urns, white ball, was it a six?

Clear? Now, remember our mystery from the start? It is Example two, a hard one. We throw a die. On a six, a ball is drawn from urn A, with two white and three black balls; otherwise from urn B, with four white and one black. The ball is white. Find the probability that the die showed six. Pause the video and try it first. Done? Okay, let's check. The die decides the urn, so the cases are E one, a six, and E two, not a six. Exactly one of them happens, so they form a partition. P of E one is one over six, and P of E two is five over six. Okay? Let W be the event that the ball is white. On a six we use urn A: two white out of five balls, so P of W given E one is two over five. Otherwise we use urn B: four white out of five, so P of W given E two is four over five. Right? Now, the ball is white, and urn A gives white with chance two over five. So the answer is two over five, isn't it? Not at all, and it is the same trap as before. Two over five is P of W given E one, but we want P of E one given W, the other way round. So first, total probability; the dot means times. One over six times two over five gives two over thirty, and five over six times four over five gives twenty over thirty. So P of W is twenty two over thirty. With me? Now Bayes: the six branch over the total. Two over thirty, divided by twenty two over thirty; the slash is another way to write over. The thirties cancel, and two over twenty two gives one over eleven. Got it? So, back to our mystery. Before the ball, a six had chance one over six. After seeing white, it is one over eleven, smaller, as the less-than sign shows. It went down, because urn A has fewer white balls. Clear?

7:09Quick revision: flip a condition, Bayes, check conditions

So, quick revision. First, to flip a condition, go through P of A intersection B; never just swap. Second, Bayes' theorem is your branch over the sum of all the branches ending in A, okay? And third, check the conditions first: P of B greater than zero, a partition, and P of A greater than zero. Now go through the other problems of Set two on this page, and try the practice problem. All the best!