SOLUTION: In a bag of M&M’s there are 12 green, 7 blue, 11 brown, 9 orange, and 15 red candies. If one candy is chosen from this class at random, what is the probability that the selected

Algebra ->  Average -> SOLUTION: In a bag of M&M’s there are 12 green, 7 blue, 11 brown, 9 orange, and 15 red candies. If one candy is chosen from this class at random, what is the probability that the selected       Log On


   

Question 648591: In a bag of M&M’s there are 12 green, 7 blue, 11 brown, 9 orange, and 15 red candies. If one candy is chosen from this class at random, what is the probability that the selected candy will be green or orange?
Found 2 solutions by MathGeek87, swincher4391:
Answer by MathGeek87(9) About Me  (Show Source):
You can put this solution on YOUR website!
12/54+9/54=21/54

Answer by swincher4391(1107) About Me  (Show Source):
You can put this solution on YOUR website!
We have 12+7+11+9+15 M&Ms. This is 54 M&Ms. This is our sample space: S. Let E be the event that you pick a green or orange M&M. There are 12+9 Green or Orange M&Ms. That means 21 satisfy this event. Then P[green or orange m&m] = N(E)/N(S) = 21/54 where N(E) is just the number of elements in that set: 21.
I know that was formal so let me break it down.
You have 54 M&Ms to choose from. 21 are green or orange. When you reach in the bag, you have a 21/54 chance of picking a green or an orange.