SOLUTION: A candy bowl contains 8 red m&ms 5 blue 3 yellow if two m&ms are selected at random what is the probability that one will be blue or one will be yellow? 3/16+5/16= 8/16= 50%?

Algebra ->  Probability-and-statistics -> SOLUTION: A candy bowl contains 8 red m&ms 5 blue 3 yellow if two m&ms are selected at random what is the probability that one will be blue or one will be yellow? 3/16+5/16= 8/16= 50%?      Log On


   



Question 327784: A candy bowl contains 8 red m&ms 5 blue 3 yellow if two m&ms are selected at random what is the probability that one will be blue or one will be yellow?

3/16+5/16= 8/16= 50%?

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A candy bowl contains 8 red m&ms 5 blue 3 yellow if two m&ms are selected at random what is the probability that one will be blue or one will be yellow?
3/16+5/16= 8/16= 50%?
Don't add them, multiply them.
If you add them, the probability increases with added restrictions, which makes no sense.
And after getting the 1st one, only 15 are left.
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The yellow is 3/16
The blue is then 5/15
--> (3/16)*(5/15) = 1/16