Selections without replacement, tree diagrams and conditional probability

Last time

Before we start: last time, in Probability, you learned these points. Here they are again on one card. Keep them in mind, because today builds on them.

One after another

When items are chosen one after another and not put back, the probabilities change at each step. This lesson covers selections without replacement, tree diagrams and conditional probability, and independent events in real situations.

Without replacement

A bag holds 5 red and 3 blue counters. Two are taken out, without replacement. P(first red) is 5 over 8. If it was red, 4 red remain out of 7, so P(second red) is 4 over 7. P(both red) is 5 over 8 times 4 over 7, which is 20 over 56, or 5 over 14.

Tree diagrams

A tree diagram shows each pick as a set of branches, with a probability on each branch. Multiply along a path for the probability of that sequence; add the paths that give the event you want. P(one of each) is red then blue, plus blue then red: 15 over 56 plus 15 over 56, which is 30 over 56, or 15 over 28.

All four paths

The four paths of the tree and their probabilities.

Checking and at least one

20 plus 15 plus 15 plus 6 is 56, so the probabilities add up to 1: a useful check. P(at least one blue) is 1 minus P(no blue), which is 1 minus 20 over 56: 36 over 56. This is quicker, and safer, than adding every path that contains a blue.

Conditional probability

The probability of B given A is P(A and B) divided by P(A). Given that both counters are the same colour, the probability that both are red is 20 over 56, divided by 26 over 56. That is 20 over 26, which is 10 over 13.

Listing selections

To choose 2 players from 4, A, B, C and D, list the pairs: AB, AC, AD, BC, BD, CD. There are 6. If every pair is equally likely, P(A is chosen) is 3 over 6, which is one half. A systematic list, keeping the letters in order, avoids counting AB and BA as different.

Independent events in context

Two machines each break down on a given day with probability 0.1, independently. P(both break down) is 0.1 times 0.1, which is 0.01; P(at least one works) is 1 minus 0.01, which is 0.99. In context, ask whether independence is realistic: machines that share a power supply may break down together.

Mistakes that lose marks

Mistake one: keeping the same total for the second pick when there is no replacement. Mistake two: forgetting that one of each can happen in two orders. Mistake three: for at least one, adding many paths and missing some, instead of 1 minus P(none). The fix: change the totals, count both orders, and use 1 minus.

The full-marks tip

The full-marks tip: write each branch probability as a fraction over the new total, and show the multiplication. For a given-that question, write P(A and B) over P(A) before any numbers.

On one card

So, on one card: selections and conditional probability. Keep it for revision.