Question
In: Math
A fun size bag of M&Ms has 4 blue, 3 orange, 3 red, 2 green, 2...
A fun size bag of M&Ms has 4 blue, 3 orange, 3 red, 2 green, 2 yellow, and 1 brown M&Ms. What is the probability of randomly selecting 5 M&Ms where 3 are blue and 2 are orange?
Solutions
Expert Solution
Number of blue M&Ms in a fun size bag = 4
Number of orange M&Ms in a fun size bag = 3
Number of red M&Ms in a fun size bag = 3
Number of green M&Ms in a fun size bag = 2
Number of yellow M&Ms in a fun size bag = 2
Number of brown M&Ms in a fun size bag = 1
Total number of M&Ms in a fun size bag = 4+3+3+2+2+1=15
Probability of randomly selecting 5 M&Ms where 3 are blue and 2 are orange
= (Number of ways of selecting 3 blue M&Ms from 4 blue M&Ms x Number of ways of selecting 2 orange M&Ms from 3) / Number of ways of selecting 5 M&Ms from a total of 15 M&Ms
Number of ways of selecting 3 blue M&Ms from 4 blue M&Ms
= 
=4
Number of ways of selecting 2 orange M&Ms from 3 = 
=3
Number of ways of selecting 5 M&Ms from a total of 15
M&Ms=
=3003
Probability of randomly selecting 5 M&Ms where 3 are blue and 2 are orange
= (Number of ways of selecting 3 blue M&Ms from 4 blue M&Ms x Number of ways of selecting 2 orange M&Ms from 3) / Number of ways of selecting 5 M&Ms from a total of 15 M&Ms
= (4 x 3)/3003 = 12/3003 = 0.003996004
Probability of randomly selecting 5 M&Ms where 3 are blue and 2 are orange = 0.003996004
milcah answered 2 months agoRelated Solutions
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Blue
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Green
Yellow
Red
Brown
Total Number of Candies
1
9
13
14
10
7
7
60
2
13
10
6
9
9
8
55
3
13
12
4
10
9
6
54
4
16
13
8
6
6
8
57
5
10
10
12
5
15
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56
6
9
18
3
6
12
12
60
7
11
13
6
15
8
6
59
8
12
18
5
9
6
5
55
9
12
10
8
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9
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12
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10
9
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10
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18
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6
12
12
60
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11
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