Question

- 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 red?

Answer #1

Given in the question

Fun size bag has 4 blue, 3 orange, 3 red, 2 green, 2 yellow and 1
brown M&Ms

So total No. of M&Ms are = 4+3+3+2+2+1 = 15

No. of randomly selected M&Ms are 5

Out of 5 M&Ms, 3 are blue and 2 are red so

Total no. of ways 3 blue M&Ms out of 4 = 4C3 = 4

Total no. of ways 2 red M&Ms out of 3 = 3C2 = 3

So total no. of ways 3 are blue and 2 are red = Total no. of ways 3
blue M&Ms out of 4 * Total no. of ways 2 red M&Ms out of 3
= 4*3 = 12

Total no. of ways 5 M&Ms out of toal 15 = 15C5 =
(15!/(5!*(15-5)!) = 3003

probability of randomly selecting 5 M&Ms where 3 are blue and 2
are red = (Total no. of ways 3 blue M&Ms out of 4 * Total no.
of ways 2 red M&Ms out of 3)/Total no. of ways 5 M&Ms out
of toal 15 = 4*3/3003 = 12/3003 = 0.004

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Red
Orange
Yellow
Green
Blue
Brown
2
1
2
3
5
1
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P(Yellow) ______________
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20%
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Blue
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Yellow
20
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17
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14
15
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3
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2
2
3
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1
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2
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5
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