*Ahem* Note what I have redded and bolded ....
(7/10)(7/10) + (3/10)(3/10) = 49/100 + 9/100 = 29/50.The Problem: Here is a game you might play. An urn contains seven red and three green balls. You are to select a ball, note its colour, and then replace it. Your oponent us then to select a ball. Ry is the event that you select red:Ro us the event that your oponent selects red;Gy is the event that you select green, etc. You win if the ball your opponent selects has the same colour as the ball you selected. What is the probability of winning? Make a tree diagram of possible outcomes.
Branch I leads to Ry
Branch 2 leads to Gy
Branch I branches in two directions
Branch III leads to Ro
Branch IV leads to Go
Branch II branches in two directions
Branch V leads to Go
Branch VI leads to Ro
To find the probability of winning, I simply add the probabilities of each event which leads to a win.
The probability of you getting two reds is (7/10)(6/9).
The probability of you getting two greens is (3/10)(2/9).
Both cases lead to a win.
So I added the two cases.
P(winning) = (7/10)(6/9) + (3/10)(2/9)
= 42/90 + 6/90
But, so the key informs me, the answer is 29/50. How could this be?