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.
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----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---WhiteI've been taught you multiply across for the probability.---6/21--- Red
---7/21---Yellow
---3/21---Green
---5/21---White
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.
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: . Right!
However, I do not know why my tree diagram isn't working.
You made it complicated, but it is correct.
I've been taught you multiply across for the probability. .Correct!
I get two values for the probability.
. .
Do you not add these two together to get ? . Yes!