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Probability · Combined events & tree diagrams

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Tree diagrams & independent events

Building a tree from a worded scenario, and why you multiply along a branch and add between branches — plus the branches-sum-to-1 check and the "at least one" complement that turns seven calculations into one.

Probability · Combined events & tree diagrams

Tree diagrams & independent events

Building a tree from a worded scenario, and why you multiply along a branch and add between branches — plus the branches-sum-to-1 check and the "at least one" complement that turns seven calculations into one.

Why it works

A two-stage experiment is just this happens, then that happens. A tree diagram is the picture of exactly that: one set of branches for everything stage 1 could do, and then, growing out of the end of every one of those, a fresh set of branches for everything stage 2 could do. Follow a branch from the start to a tip and you have traced one complete outcome of the whole experiment. Nothing about the diagram is decorative — every number on it earns its place.

Everything after that is two rules, multiply along, add across, and both are consequences of what a probability is, not conventions to be memorised.

Independent, and why "replaced" is the magic word. Two events are independent when knowing one happened tells you nothing about the chances of the other. Rolling a dice twice: independent. Spinning a spinner and tossing a coin: independent. Taking a counter from a bag, putting it back, and taking another: independent, because replacing it restores the bag exactly as it was, so the second draw meets an identical situation. (Take a counter and keep it and the second set of branches carries different numbers — that is a different concept.) The practical meaning is simple: when the events are independent, every second-stage branch carries the same probabilities no matter which first-stage branch you came down.

Why you multiply along a branch. Imagine running the experiment 10001000 times. Suppose stage 1 gives red with probability 0.30.3. Then about 300300 of those 10001000 runs come down the red branch. Those 300300 runs now face stage 2, where a head has probability 0.60.6; independence says they behave like any other runs, so about 0.6×300=1800.6 \times 300 = 180 of them give a head. That is 180180 of the original 10001000 runs which went red and then head:

P(red, then head)=1801000=0.18=0.3×0.6.P(\text{red, then head}) = \frac{180}{1000} = 0.18 = 0.3 \times 0.6.

Multiplying is simply what "a share of a share" does. Six tenths of three tenths is 610×310=18100\frac{6}{10} \times \frac{3}{10} = \frac{18}{100}. Or picture a square of area 11: cut a strip of width 0.30.3, then cut 0.60.6 of the height of that strip, and the rectangle left has area 0.3×0.60.3 \times 0.6. Each branch narrows the field; narrowing twice multiplies.

This is also why adding cannot be right for "both". Probabilities are never more than 11, so multiplying makes things smaller — and demanding that two things both happen is harder than demanding just one of them, so the answer must come out smaller than either probability on its own. Write 0.3+0.6=0.90.3 + 0.6 = 0.9 for "both" and you have claimed that needing two things is more likely than needing only the head (0.60.6). With 0.6+0.70.6 + 0.7 you would get 1.31.3, which is not even a probability.

Why you add between branches. Two different complete paths through the tree can never both happen on one run — the experiment finishes at exactly one tip. Outcomes that cannot happen together are mutually exclusive, and mutually exclusive shares of the whole simply add, with nothing counted twice. So "exactly one head in two tosses" means the path head-then-tail or the path tail-then-head, and

P(exactly one head)=P(HT)+P(TH).P(\text{exactly one head}) = P(HT) + P(TH).

The word in the question is the tell: AND along a path (multiply), OR between paths (add). Hold on to the reason, though, not the slogan — you multiply because each branch cuts down what is left, and you add because the paths are alternatives.

Why every set of branches from one node sums to 1. At a node the experiment is about to do exactly one thing, and the branches list every possibility, each ruling out the others. Together they are the whole of what can happen, so their probabilities total 11. That hands you any missing number for free: if a spinner has P(red)=0.5P(\text{red}) = 0.5, P(blue)=0.2P(\text{blue}) = 0.2 and the only other colour is green, then P(green)=1−0.5−0.2=0.3P(\text{green}) = 1 - 0.5 - 0.2 = 0.3. It is a free check too — work out all the tip probabilities and they must also total 11.

"At least one" — the trick worth the most marks. "At least one head in three tosses" means one head, or two, or three: seven separate paths to find and add. But its opposite is a single path — no heads at all. Between them "at least one" and "none" cover everything and overlap nowhere, so

P(at least one)=1−P(none).P(\text{at least one}) = 1 - P(\text{none}).

For three tosses of a fair coin, P(none)=0.53=0.125P(\text{none}) = 0.5^3 = 0.125, so P(at least one)=1−0.125=0.875P(\text{at least one}) = 1 - 0.125 = 0.875. One line instead of seven.

The classic wrong move is to add the separate probabilities: "the bus is late with probability 0.20.2 each day, so over two days it is 0.2+0.2=0.40.2 + 0.2 = 0.4." Run that method for five days and it gives 1.01.0 — certain — and for six days it gives 1.21.2, which is impossible. It breaks because it double-counts the runs where the bus is late on both days. The correct answer for two days is 1−0.8×0.8=1−0.64=0.361 - 0.8 \times 0.8 = 1 - 0.64 = 0.36, which is a little less than 0.40.4, exactly the amount that was double-counted (0.2×0.2=0.040.2 \times 0.2 = 0.04). ✓ The complement never has that problem, because "none" is a single path.

Three stages. Nothing new. Three sets of branches, three probabilities multiplied along each path, alternatives added at the end. For three independent trials each with probability pp of success, P(all three)=p3P(\text{all three}) = p^3 and P(at least one)=1−(1−p)3P(\text{at least one}) = 1 - (1-p)^3.