Branching Out

Branching Out

When two things happen in a row, two coin flips or two counters drawn, a probability tree lays out every combined outcome so nothing is missed. Two rules run the whole topic: multiply along a path, and add between the paths that satisfy your event. The second of those is the one most answers forget, so this module makes you use it.

The words for it

Five terms. The two rules in the last one are the whole method, and knowing which applies when is most of the skill:

Read the tree

This tree shows two flips of a fair coin. Tap the outcome HH (Heads on the first flip AND Heads on the second).

Heads then heads

A fair coin is flipped twice. What is the probability of getting Heads on the first flip and Heads on the second? Give a decimal.

Multiply along, add between

One combined outcome needs one path, so you multiply along it. But most questions describe an event that several paths satisfy, and then you add. Take two coin flips and ask for exactly one head. That is not one outcome, it is two: HT and TH. • Multiply along the HT path: 0.5 × 0.5 = 0.25 • Multiply along the TH path: 0.5 × 0.5 = 0.25 • Add between them: 0.25 + 0.25 = 0.5 Missing the second path is the single most common way to lose these marks. Whenever a question says "exactly one", "at least one" or "one of each", expect more than one path.

Exactly one win

A spinner lands on "win" with probability 0.2 each spin, and it is spun twice. What is the probability of winning EXACTLY ONE of the two spins? Give a decimal.

Multiply or add?

A bag gives a red counter with probability 0.4. Two counters are drawn with replacement, and you want the probability of getting AT LEAST ONE red. Which operations does that need?

• Both: multiply along each of the three paths that contain a red, then add those three results together
• Multiplication only, since 0.4 × 0.4 gives the chance of drawing red
• Addition only: 0.4 + 0.4 gives the chance across two draws
• Subtraction only, taking 0.4 away from 1

Does it go back in?

A bag holds 3 red and 2 blue counters, and two are drawn. Whether the first is replaced changes the second probability, and it changes it in two ways at once:

Two reds

A bag holds 3 red and 2 blue counters. Two are taken WITHOUT replacement. What is the probability that both are red? Give a decimal.

At least one

P(at least one) = 1 − P(none). For two coin flips that saves a little work. For anything larger it is the difference between one line and a page. Roll a fair die four times and ask for the probability of at least one six. Done directly, you would have to find the chance of exactly one six, exactly two, exactly three and exactly four, and add all four results, each of which needs several paths. Done through the opposite: the chance of no six on one roll is 5/6, so on four rolls it is (5/6)⁴, which is about 0.482. Therefore P(at least one six) is about 1 − 0.482 = 0.518, a little better than an even chance. One subtraction, and you are finished.

Why take it from 1?

Why is P(at least one) = 1 − P(none) a valid shortcut rather than an approximation?

• Because "at least one" and "none" between them cover every possible result and cannot both happen, so their probabilities must add to exactly 1
• Because P(none) is usually small, so subtracting it from 1 gives a close answer
• Because the events are independent, and independent probabilities always sum to 1
• Because subtraction is a simpler operation than addition, so it introduces less error

Which are true?

Select ALL THREE statements that are TRUE.

• A question asking for "exactly one" of something needs both rules: multiply along each qualifying path, then add between them
• Drawing without replacement changes both the number of that colour left AND the total number of counters
• The branches leaving any single point must add to 1, because between them they cover every possibility at that stage
• You add the probabilities along a single path
• A single branch can carry a probability greater than 1 if the event is very likely
• Removing a counter without replacing it changes the total number of counters but not the number of that colour

The tree rules

On a probability tree, _____ the probabilities along a path to find one combined outcome, and _____ the results of the different paths whenever more than one outcome counts as your event. The branches from any single point always add up to _____. For "at least one", it is quicker to find the probability of _____ and subtract it from 1.