SOLUTION: In a particular bag of of skittles, there are 13 red, 12 orange, 10 green, 12 yellow, and 11 purple skittles.
Answer as a fraction or w/o rounding
1. Selecting a yellow skittle _
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Question 1182296: In a particular bag of of skittles, there are 13 red, 12 orange, 10 green, 12 yellow, and 11 purple skittles.
Answer as a fraction or w/o rounding
1. Selecting a yellow skittle _______
2. Selecting a red or purple skittle ______
3. Not selecting a green skittle ______
4. Selecting two consecutive orange skittles _____
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
Problem 1
There are 12 yellow out of 13+12+10+12+11 = 58 total.
This forms the fraction 12/58 = (2*6)/(2*29) = 6/29 which is the probality of selecting a yellow skittle.
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Problem 2
We have 13 red + 11 purple = 24 skittles we want to select out of 58 total
24/58 = (2*12)/(2*29) = 12/29 is the probability of selecting either a red or purple skittle (note the events are mutually exclusive).
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Problem 3
There are 10 green and 58-10 = 48 non-green skittles.
48/58 = (2*24)/(2*29) = 24/29 is the probability of selecting a skittle that isn't green.
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Problem 4
This is assuming you aren't putting the first one back, or no replacement has been made.
12/58 represents the probability of selecting an orange skittle (12 orange out of 58 total)
11/57 is the probability of selecting another orange skittle (12-1 = 11 orange left, out of 58-1 = 57 total left)
Multiply the probabilities
(12/58)*(11/57)
(12*11)/(58*57)
132/3306
22/551 is the probability of selecting two orange skittles in a row (no replacement).
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