Probability Tree Diagrams and Independent Events 1

Grace removes one ball from the bag, writes down the colour and replaces the ball back into the bag. Grace then removes another ball from the bag.

In the first round, Grace can choose either a red ball or a yellow ball. Therefore, we can display these outcomes as 'branches'.

There are 6 red and 7 yellow balls in the bag. $6+7=13$, so there are 13 balls in total. We can see that 7 out of 13 balls are yellow, so we can assign it the probability $\dfrac{7}{13}$.

We can display these probabilities on the corresponding branches.

Grace replaces the ball, and takes another ball out of the bag.

Now our probability tree is complete, we can work out the combined probability of different ball combinations.

There are two ways this can happen. Either Grace can pick two red balls in a row, or two yellow balls in a row.

We can find this by applying the "and" rule. $\dfrac{6}{13} \times \dfrac{6}{13}=\dfrac{36}{169}$

Each time, yellow has the probability $\dfrac{7}{13}$. So the probability of both being yellow is $\dfrac{7}{13} \times \dfrac{7}{13}=\dfrac{49}{169}$

As there are two ways to pick the same colour twice in a row, we need to add the probabilities together.

We can write this as P(same colours)=$\dfrac{85}{169}$

This could happen if Grace picks a red ball first and then a yellow ball OR if Grace picks a yellow ball first and then a red ball. Note that the order is very important, as they are two different scenarios!

We can find this by multiplying the probabilities together. $\dfrac{7}{13} \times \dfrac{6}{13}=\dfrac{42}{169}$

We can find this by multiplying the probabilities together. $\dfrac{6}{13} \times \dfrac{7}{13}=\dfrac{42}{169}$

As there are two ways to pick different coloured balls across the two events, we can add the probabilities together.

We can write this probability as P(Different Colours)=$\dfrac{84}{169}$

There are three possibilities to include here. Our calculation could therefore be $P(Yellow,Yellow) + P(Yellow,Red) + P(Red,Yellow)$

The only way to not pick a yellow ball at least one of the times is to pick red both times. Therefore, we can find the probability of picking red both times, and subtract this from the total probability.

We can take this value away from the total probability, $1$, to find the probability of picking at least one yellow.

Nice!