# Math Help - Confused about probability and Tree diagrams

1. ## Confused about probability and Tree diagrams

The problem is
A jar contains 6 red balls, 3 green balls, 5 white balls and 7 yellow balls. Two balls are chosen from the jar, with replacement. What is the probability that you choose a red and a yellow ball?

I know that they do not affect each other so could be independent events, so P(red and yellow) = P(red) x P(yellow) so 6/21 x 1/3 = 2/21

However, I do not know why my tree diagram isn't working

----6/21---- Red
---6/21--- Red
---7/21---Yellow
---3/21---Green
---5/21---White
----7/21---- Yellow
---6/21--- Red
---7/21---Yellow
---3/21---Green
---5/21---White
----3/21---- Green
---6/21--- Red
---7/21---Yellow
---3/21---Green
---5/21---White
----5/21---- White
---6/21--- Red
---7/21---Yellow
---3/21---Green
---5/21---White
I've been taught you multiply across for the probability.
I get two values for the probability.
P(yellow and red)=2/21
P(red and yellow)=2/21

Do you not add these two together to get 4/21?

I'm really confused now as to which one is correct.

2. The order matters as replacement is allowed.

Choosing a red and a yellow can happen in 2 ways....

First a red is chosen, then a yellow.
Or, first a yellow is chosen, then a red.

P(red and yellow) is just one of these.

3. Hi Henryblah,
P of red or yellow first draw = 13/21
P of red or yellow not being drawn on second draw = 9/21
P of red and yellow two draws = 13/21 - 9/21 = 4/21

4. Hello, Henryblah!

A jar contains 6 red balls, 3 green balls, 5 white balls and 7 yellow balls.
Two balls are chosen from the jar, with replacement.
What is the probability that you choose a red and a yellow ball?
I assume this means "in any order?".

I know that they do not affect each other so they are independent events.
So: $P(\text{red \& yellow}) \,=\, P(\text{red}) \times P(\text{yellow}) \,=\,\frac{6}{21} \times \frac{1}{3} \:=\: \frac{2}{21}$ . Right!

However, I do not know why my tree diagram isn't working.
You made it complicated, but it is correct.

$\begin{array}{ccc}\text{Red: } \frac{6}{21} \\
& \text{Yellow:} & \frac{7}{21} \\ \\[-4mm]
& \text{Other:} & \frac{14}{21} \end{array}$

$\begin{array}{ccc}\text{Yellow: }\:\frac{7}{21} \\
& \text{Red:} & \frac{6}{21} \\ \\[-4mm]
& \text{Other:} & \frac{15}{21} \end{array}$

I've been taught you multiply across for the probability. .Correct!

I get two values for the probability.
. . $\begin{array}{c}P(\text{Yellow, then Red})\,=\,\frac{2}{21} \\ \\[-4mm]
P(\text{Red, then Yellow})\,=\,\frac{2}{21} \end{array}$

Do you not add these two together to get $\dfrac{4}{21}$? . Yes!

5. Thank you. I think I understand this now.

Also @ whoever put my thread here
sorry for putting this in the wrong section. I didn't see there was a probability forum.