Question 1196701
<font color=black size=3>
spinner A:  1/2 green, 1/4 red, 1/4 blue
spinner B: 2/3 red, 1/6 green, 1/6 yellow


Note: if spinner B has half red on one side, and a 1/6 slice of red on the other side, then 1/2+1/6 = 3/6+1/6 = 4/6 = 2/3 is how much of a fraction red takes up in total.


P(2 green) = P(A is green)*P(B is green)
P(2 green) = (1/2)*(1/6)
P(2 green) = 1/12


P(2 red) = P(A is red)*P(B is red)
P(2 red) = (1/4)*(2/3)
P(2 red) = 2/12



P(2 blue) = P(A is blue)*P(B is blue)
P(2 blue) = (1/4)*(0)
P(2 blue) = 0
It is impossible to get 2 blue since spinner B does not have blue.



P(2 yellow) = P(A is yellow)*P(B is yellow)
P(2 yellow) = (0)*(2/3)
P(2 yellow) = 0
It is impossible to get 2 yellow since spinner A does not have yellow.


So either we get 2 red or 2 green if we want two of the same color.
P(2 same color) = P(2 red OR 2 green) 
P(2 same color)=  P(2 red) + P(2 green)
P(2 same color) = 2/12 + 1/12
P(2 same color) = (2 + 1)/12
P(2 same color) = 3/12
P(2 same color) = 1/4
Note: The events "2 red" and "2 green" are mutually exclusive. 


Then we can say
P(2 red, given 2 of the same color) = P(2 red)/P(2 same color)
P(2 red, given 2 of the same color) = (2/12)/(1/4)
P(2 red, given 2 of the same color) = (2/12)*(4/1)
P(2 red, given 2 of the same color) = 8/12
P(2 red, given 2 of the same color) = 2/3


And,
P(2 green, given 2 of the same color) = P(2 green)/P(2 same color)
P(2 green, given 2 of the same color) = (1/12)/(1/4)
P(2 green, given 2 of the same color) = (1/12)*(4/1)
P(2 green, given 2 of the same color) = 4/12
P(2 green, given 2 of the same color) = 1/3


The results 2/3 and 1/3 are complementary to represent all the possible ways to get 2 of the same color. For more information, check out the concept of conditional probability.



------------------------------------------------------------

Another approach:


We can use table to lay out the possible outcomes.
Along the top row are the four slices of green, green, red and blue for spinner A.
The outcomes for spinner B are along the left hand side.<table border = "1" cellpadding = "5"><tr><td></td><td></td><td>1</td><td>2</td><td>3</td><td>4</td></tr><tr><td></td><td></td><td>green</td><td>green</td><td>red</td><td>blue</td></tr><tr><td>1</td><td>red</td><td></td><td></td><td>X</td><td></td></tr><tr><td>2</td><td>red</td><td></td><td></td><td>X</td><td></td></tr><tr><td>3</td><td>red</td><td></td><td></td><td>X</td><td></td></tr><tr><td>4</td><td>red</td><td></td><td></td><td>X</td><td></td></tr><tr><td>5</td><td>red</td><td></td><td></td><td>X</td><td></td></tr><tr><td>6</td><td>red</td><td></td><td></td><td>X</td><td></td></tr><tr><td>7</td><td>green</td><td>X</td><td>X</td><td></td><td></td></tr><tr><td>8</td><td>green</td><td>X</td><td>X</td><td></td><td></td></tr><tr><td>9</td><td>red</td><td></td><td></td><td>X</td><td></td></tr><tr><td>10</td><td>red</td><td></td><td></td><td>X</td><td></td></tr><tr><td>11</td><td>yellow</td><td></td><td></td><td></td><td></td></tr><tr><td>12</td><td>yellow</td><td></td><td></td><td></td><td></td></tr></table>
Each X represents a situation where we get 2 of the same color. Either 2 red or 2 green. The cells left blank are when we get a mix of two different colors, so we'll ignore those cases. The numbers along the left and top are there to help keep track of the slices.


There are 12 X's total.


Four of the Xs are when we get 2 greens
4/12 = 1/3
while the remaining eight are 2 reds
8/12 = 2/3


------------------------------------------------------------

Answers: 


If we get 2 of the same color, then,
probability of 2 reds = 2/3
probability of 2 greens = 1/3
</font>