Probability can be presented using **tree diagrams**. Each branch of the tree represents an outcome (similar to a frequency tree diagram, but each branch is labelled with a probability, not a frequency).

All outcomes must be shown from each node. For example, a bag of balls contains 4 red balls and 6 blue balls. **P**(red) = `frac(2)(5)` and **P**(blue) = `frac(3)(5)`:

If, in the above example, a ball is drawn, replaced and then a second ball is drawn:

It is important to determine whether, on a subsequent event, whether this depends on the result of the previous event.

Multiply the probabilities *along* branches to calculate the probability of two consecutive events. The probability of drawing two red balls is **P**(red) x **P**(red) = `frac(2)(5)` x `frac(2)(5)` = `frac(4)(25)`.

If the balls were NOT replaced, then the denominator on the second event as there are fewer bags in the ball to draw from. This gives a different answer:

Adding the ends of all the branches gives a probability of 1.

A bag contains 12 balls. 3 are red and 9 are blue. A ball is drawn at random, and replaced in the bag. A second ball is drawn at random. What is the probability of drawing 2 red balls?

Draw the probability tree. For two red balls, multiply along the tree:

**P**(red) x **P**(red) = `frac(3)(12)` x `frac(3)(12)` = `frac(9)(144)` = `frac(1)(16)`

Answer: `frac(1)(16)`

A bag contains 12 balls. 3 are red and 9 are blue. A ball is drawn at random, and NOT replaced in the bag. By drawing a probability tree, or otherwise, show that the probability of drawing two consecutive balls of the same colour is `frac(21)(33)`.

Note that the balls are NOT replaced: therefore the denominator of the fractions changes between the first draw and the second draw.

Answer: Probability of two reds or two blues:

(**P**(red_{1}) x **P(red)**_{2}) + (**P**(blue)_{1} x **P**(blue)_{2})

`frac(3)(12)` x `frac(2)(11)` + `frac(9)(12)` x `frac(8)(11)` = `frac(13)(22)`

See also Independent and Dependent Events

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