p(RYR)= 3/5 x 2/4 X 2/3 = 2/10Shouldn't iii) be 0.4?
You multiply the branches of of the tree to get the probability, then add them all together..
You have the option of RYR or YRY
P(RYR) = 3/10
P(YRY) = 1/10
Therefore total probability = 4/10
Checked maths textbook to make sure and that is what they do and I think it is the correct method :\
No.Shouldn't iii) be 0.4?
You multiply the branches of of the tree to get the probability, then add them all together..
You have the option of RYR or YRY
P(RYR) = 3/10
P(YRY) = 1/10
Therefore total probability = 4/10
Checked maths textbook to make sure and that is what they do and I think it is the correct method :\
Fuck. Fingers crossed I wrote the tree diagram right in the exam >.<p(RYR)= 3/5 x 2/4 X 2/3 = 2/10
No.
P(R, Y, R) = (3/5).(1/2).(2/3) = 6/30 = 1/5
P(Y, R, Y) = (2/5).(3/4).(1/3) = 6/60 = 1/10
1/5 + 1/10 = 3/10
yeh l did taht aswellfor the third part ab out consecutive days.. I did
P(YRY) = 2/5 x 3/4 x 1/3
P (RYR) = 3/5x2/4x2/3
and added P(YRY)+ P(RYR) together
I think it's right