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Probability tree diagrams

Let’s see how we can represent the situation of taking of taking balls out of a bag with a diagram.

The bag contains 4 red balls and 7 green balls.  We are going to take out two balls one at a time.  We will not be putting the balls back in the bag.

The diagram

This diagram shows what could happen at each stage of the trial, but so far we have not introduced any probabilities.  We know that for the first ball P(red)=\frac{4}{11}\mbox{ and }P(green)=\frac{7}{11}.  We place these values in the appropriate place.

After we have picked out a red ball we are left with a bag with 3 red balls and 7 green balls.

After we have picked out a green ball we are left with a bag with 4 red balls and 6 green balls.

We can calculate the appropriate probabilities and put them in the right place.

It is always useful to work out the values at the end.  The end of red, red means the probability of getting red AND then another red.  We learnt earlier that ‘AND’ means \times, so to work out the ends of the branches we multiply the probabilities.

Questions

P(red, red)

Here we just the end of that branch.  Therefore the answer is \frac{12}{110}

P(red and blue)

This could be P(red, blue) or P(blue, red).  ‘OR’ means add so we add the ends of those two branches.  Therefore the answer is \frac{56}{110}

P(not 2 greens)

Here we want 1-P(green,green)=\frac{68}{110}

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