Question 1174432: In a complete graph with 12 nodes you choose the color for the edges in a following manner: You toss a coin for each edge. If the coin shows heads, you toss a three sided fair die. If the die shows 1 you color the edge blue, if the die shows 2 you color the edge yellow and in case the die shows 3 you color the edge green. In case the coin shows tails, you toss a three sided fair die also. This time, if the die shows 1 or 2 the edge is colored yellow, while if it shows 3 the edge is colored red. What is the expected number of edges colored red or blue with this strategy? (give only the expected number of edges as an answer, no text)
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
(heads, 1) --> blue
(heads, 2) --> yellow
(heads, 3) --> green
(tails, 1) --> yellow
(tails, 2) --> yellow
(tails, 3) --> red
All outcomes have the same probability, (1/2)(1/3) = 1/6.
P(red OR blue) = 1/6+1/6 = 1/3
Probability 1/3 and 12 edges to be colored means the expected number of edges colored red or blue is 12(1/3) = 4.
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Note: I ignored your instruction "(give only the expected number of edges as an answer, no text)". Simply giving you the answer "4" wouldn't have helped you learn anything.
You might think about taking the time to look at what you are posting, rather than copying and pasting the whole assignment into your post.
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