Partitioning the sample space, and totalling a probability across the parts

0:00Two bags on a table. The first holds two white balls and three red. The second holds four white and five red. Somebody picks a bag at random, reaches in without looking, and draws one ball. You do not see which bag they used. Some questions about this are easy. Which bag did they pick? A half each, by construction. If they used the first bag, what is the chance the ball is white? Two out of five.

0:31Those are easy because they run forwards. The bag comes first, the ball comes second, and you are asking about the ball given the bag. Here is the question that is not easy. The ball on the table is white. What is the chance it came from the second bag? That runs backwards. You are being asked about the first stage given the second, and nothing you have been handed points that way.

1:01It is worth being precise about what is missing, because the repair is the whole of this topic. Write down what the setup gives you. The chance of each bag. And the chance of white given each bag. Those are the arrows that point forwards, one for each branch, and they were free. The arrow you want points the other way: the chance of a bag, given white. There is a definition that turns one into the other. The chance of the bag given white is the chance of both divided by the chance of white. The numerator you can build with the multiplication rule. So really only one thing is missing.

1:45The chance of white. Full stop. Not white from bag one, not white from bag two. White. And nobody handed you that number. It has to be assembled from the pieces, and the machine for assembling it is what the rest of this video is. Here is the idea, and it is almost too plain to state. Cut the sample space into pieces. Ask your question separately inside each piece. Then weight each answer by how likely that piece was, and add.

2:19That is it. But it only works if the cut is a proper one, and a proper cut is three demands at once. One. No two pieces share an outcome. Two. The pieces together make up the whole space. Nothing is left over. Three. Every piece carries some probability. None of them is an event that never happens. A family of events meeting all three has a name: it is a partition of the sample space.

2:49Three demands is a lot to keep in your head, and the usual reaction is to memorise them. Do not. Each one is there to buy one specific line of the proof, and once you see which, there is nothing left to remember. So take them one at a time, and break them on purpose. Remove a piece, so the cut no longer covers everything. Nothing overlaps, every piece still carries probability. Only demand two is gone.

3:22What fails is the very first line, the one that says your event is recoverable from the pieces at all. Over every cut of a die and every event of it — twelve thousand nine hundred and ninety two pairs — that line holds every single time. Drop a piece and it fails at nine thousand five hundred and seventy four of them. Now put one outcome into two pieces at once. Nothing is missed, every piece still carries probability. Only demand one is gone.

3:56What fails now is a different line: the one that turns the probability of a union into a sum. Adding is what disjointness buys, and nothing else buys it. Over the same twelve thousand nine hundred and ninety two pairs that line is perfect; with an outcome counted twice it fails at exactly half of them. And the third demand. Take a die with one face that can never come up, so a piece is allowed to carry no probability at all. Now the last line of the proof — the one that writes each term as a weight times a conditional — cannot be written down, because there is no conditional to write. Of the two hundred and three ways to cut that space, fifty two contain a dead piece, and on those the answer is not a wrong number. There is no number.

4:51Three demands, three lines, one each. That is not a coincidence and it is not a mnemonic. It is what the proof actually spends. Before the general statement, two cuts worth knowing by sight. The smallest one there is: any event, and everything else. Two pieces. They share nothing, because an event and its opposite never overlap. They cover everything, because one of them always happens. Demand three is the one to watch. It holds exactly when neither piece is impossible — which means the event must not be certain and must not be impossible either.

5:35On a die, sixty two of the sixty four events give a proper two-piece cut. The two that fail are the event that never happens and the event that always does. This is worth thirty seconds because almost every two-branch problem you will ever set up uses this cut and only this cut. Struck or not struck. Faulty or sound. From the first bag or from the second. The second one to know by sight: take two events that overlap, and read four pieces off the picture.

6:12The part in both. The part in the first only. The part in the second only. And the part in neither. Every outcome of the space lands in exactly one of those four, which is demands one and two together, and you can see it rather than check it. Demand three still has to be checked, and it is the one that can fail: if the two events happen to share nothing, the first piece is empty and the family is not a cut. On a space of three tossed coins, of the sixty five thousand five hundred and thirty six ordered pairs of events, forty thousand eight hundred and twenty four give four live pieces.

6:57Notice what this buys. Two events give you a four-way cut for free, and four pieces is often exactly the resolution a problem needs. One more thing to get out of the way, because it worries people. There is more than one way to cut a space, and the different ways do not agree about anything except the answer. Take a die, and the event that the face is four or less. Two thirds, obviously.

7:27Cut one: low half and high half. Each weighs a half. Inside the low half the event is certain — one. Inside the high half it is one in three. Half of one plus half of a third. Two thirds. Cut two: pair up one with four, two with five, three with six. Each weighs a third now, and the conditionals are one, a half and a half. A third of one plus a third of a half plus a third of a half. Two thirds again.

7:59Different number of pieces. Different weights. Different conditionals. Same answer, and it has to be, because both of them are computing the same probability by rearranging the same outcomes. So choosing a cut is a modelling decision, not a mathematical risk. Choose the one whose conditionals you can actually write down. Now the statement. Let the pieces be a partition of the sample space. Then for any event at all, the chance of that event is the chance of the first piece times the chance of the event given the first piece, plus the chance of the second piece times the chance of the event given the second, and so on through every piece.

8:50Folded up, that is a sum over the pieces of a weight times a conditional. Two things to say about it before the proof. The first is that your event does not have to sit inside one of the pieces. It usually straddles several. That is the whole point — the pieces cut across the event, and the theorem collects what each one contributes. The second is that the pieces do not have to be equally likely. Nothing anywhere in this needs that. Swept over every cut of a deliberately lopsided space, where the four outcomes carry a half, a quarter, an eighth and an eighth, the result holds at every cut and every event, without exception.

9:38Four moves, and each one spends a demand you already paid for. Move one. Write your event as itself intersected with the whole sample space. That is free, and it is only true because the pieces cover everything — that is demand two, spent. Move two. The whole space is the union of the pieces, so distribute: the event is the union of the event with the first piece, the event with the second piece, and so on.

10:11Move three. Take probabilities. Those parts share no outcomes, because the pieces share no outcomes, so the probability of the union is the plain sum of the probabilities. That is demand one, spent. Move four. Each term is the chance that the event and a piece both happen, and the multiplication rule rewrites that as the chance of the piece times the chance of the event given the piece. That needs the piece to carry probability, or the conditional does not exist. Demand three, spent.

10:49Add them up and the statement is standing there. No new idea entered the room at any point — it is the multiplication rule applied once per piece, and then a sum. Work one all the way through. A construction job. There may be a strike, with probability nought point six five. If there is a strike, the job finishes on time with probability nought point three two. If there is no strike, it finishes on time with probability nought point eight.

11:25Two pieces: strike, and no strike. That is the smallest cut, and the second weight is one minus the first — nought point three five. Branch one: nought point six five times nought point three two is nought point two zero eight. Branch two: nought point three five times nought point eight is nought point two eight. Add them. Nought point four eight eight. Now the check almost nobody is taught. The answer is a weighted average of the two conditionals, with weights that add to one. So it has to land between them. Nought point three two, then nought point four eight eight, then nought point eight. It does.

12:11That check is free and it catches the commonest error in this whole topic, which is adding the conditionals instead of weighting them. Nought point three two plus nought point eight is one point one two, and a probability cannot be one point one two. Over every cut of a die and every event, the bare sum of the conditionals runs above one at nine thousand one hundred and thirteen of the twelve thousand nine hundred and ninety two cases. It is not a rare accident. It is what that mistake does.

12:50One more, where the second stage depends on the first. An urn holds five red balls and five black. Draw one, look at it, put it back, and then add two more balls of whatever colour you drew. Now draw again. What is the chance the second ball is red? Cut on the first draw, because that is the thing that changes the urn. Red first, a half. Black first, a half.

13:19If the first was red, the urn now holds twelve balls, seven of them red. The conditional is seven twelfths. If the first was black, the urn still holds twelve, but only five are red. Five twelfths. Weight and add. A half of seven twelfths plus a half of five twelfths. The twelfths add to twelve twelfths, and half of that is one half. Exactly one half — the same as the first draw. That is not a coincidence and it is worth a sentence. The two branches are equally likely and they pull the number the same distance in opposite directions, so the pulls cancel. Adding two balls of the colour you drew helps red exactly as often as it helps black.

14:10What to keep. A partition is three demands at once: no overlap, nothing left out, and every piece carrying some probability. Not a collection of events you like the look of. Each demand buys one line of the proof. Covering everything lets you recover the event. No overlap turns a union into a sum. Positive probability lets a term be written as a weight times a conditional. If you can name which line, you will never set one of these up wrongly.

14:46The result itself: total across the pieces, weight times conditional, once per piece. Weight, and then add. Never just add. The weights are what make the answer a probability, and they are what put it between the smallest conditional and the largest, which is the fastest check you have. The pieces need not be equally likely, the event need not sit inside one of them, and more than one cut will work.

15:16And the number you just learned to build — the chance of a white ball, with no bag named — is the one that was missing at the start. It is the denominator of the reversal, and reversing is what comes next.