The first draw
A bag holds 3 red counters and 2 blue counters. One counter is taken out at random, then a second, and the first is not put back. This is called drawing without replacement.
The first draw is from all 5 counters, so it is red with probability and blue with probability . After it, only 4 counters are left in the bag.
The bag has changed
If the first counter was red, the bag now holds 2 red and 2 blue. The second counter is red with probability and blue with probability . So the second branch after red reads , not : there is one red fewer and one counter fewer.
If the first counter was blue, the bag holds 3 red and 1 blue, so the second counter is red with probability and blue with probability . The second branches are different after each first branch, because what is left depends on what was taken.
R for a red counter and B for a blue one: the first draw from 5 counters, the second from the 4 that are left. Red then red is colored: after a red, the second red has chance .
The four counters that are left
Name the counters R1, R2 and R3 for the reds and B1 and B2 for the blues. A pair of draws is a first counter and a different second counter: 5 choices for the first and 4 for the second, so 5 × 4 = 20 equally likely pairs. On a grid of first counter against second counter, the 5 cells where the same counter is drawn twice are crossed out, because a counter that has been taken cannot be taken again.
Look along the row for R1 drawn first. Four cells are left, one for each counter still in the bag, and 2 of them, R2 and R3, are red. That is the on the tree.
The first counter down the side and the second across the top, with a cross where the same counter would be drawn twice. In the R1 row, the 2 red counters of the 4 that are left are colored.
With replacement
If the first counter is put back before the second draw, the bag is the same for both draws: 3 red and 2 blue. Then the second branches read and after either first branch, and the two draws are independent.
Without replacement, the first draw changes the chances on the second, so the draws are not independent. A tree for drawing without replacement has different numbers on the second branches after each first branch.
The same bag, R for red and B for blue, with the first counter put back: the second branches are and after both first branches.
Checking the second branches
Every second branch without replacement is out of one fewer counter: 4 here, not 5. The color that was drawn first has one counter fewer on top, and the other color keeps its count. The branches from each point still add to 1: and .
A bag holds 4 red and 3 blue counters, and a red is taken out. That leaves 3 red among 6 counters, so the next counter is red with probability .
The usual mistakes
Using the chance before the draw, , on the second branch. That is the bag with the red put back, but the bag now holds 6 counters, and only 3 are red.
Taking one off the top only, . The red counter has left the bag, so it comes off the number of counters too: .
Changing the second branches the same way after every first branch. After a red the reds go down by one, and after a blue the blues do.
Socks in the dark
In the application below, two socks are taken from a drawer of 5 black and 3 gray without the first being put back. The second branches are worked from the 7 socks that are left, and the paths are multiplied and added for the same color and for one of each.
Worked example: Two Socks Taken from a Drawer in the Dark, Without Replacement
Question A drawer holds 5 black socks and 3 gray socks. In the dark, Wei Ming takes out one sock and then a second, without putting the first back. (a) Find the probability that the two socks are the same color. (b) Find the probability that he takes one sock of each color.
1.The first sock is black with probability 58 and gray with probability 38.
The first sock is black with probability 58 and gray with probability 38. 2.After a black sock, 4 black and 3 gray socks are left, so the second sock is black with probability 47 and gray with probability 37. After a gray sock, 5 black and 2 gray socks are left, so the probabilities are 57 and 27.
Only 7 socks are left, and the second branches change with the first sock: 47 and 37 after a black, 57 and 27 after a gray. 3.Multiply along each path: black then black 58 × 47 = 2056, black then gray 1556, gray then black 1556, and gray then gray 38 × 27 = 656. Check: 20 + 15 + 15 + 6 = 56.
Multiply along each path. The four paths add up to 5656 = 1. 4.(a) The same color is black then black or gray then gray: 2056 + 656 = 2656 = 1328.
(a) The same color is the first path or the last: 2656 = 1328. 5.(b) One of each color is the two middle paths: 1556 + 1556 = 3056 = 1528. Check: 1328 + 1528 = 1, because the two socks are either the same color or not.
(b) One of each is the two middle paths: 3056 = 1528.
Answer: (a) 1328; (b) 1528
Common mistakes
- Using 58 and 38 again on the second branches. That describes putting the first sock back. Without replacement there are only 7 socks for the second draw, and the color of the first sock has one sock fewer.
- Counting only black then gray for part (b), which gives 1556. Gray then black is a different path that also gives one sock of each color, so both paths must be added.
More probability with several events problems, worked step by step →