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Kurt, a painter, has 9 jars of paint 4 of which are yellow, [#permalink]

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11 Sep 2007, 22:50

Kurt, a painter, has 9 jars of paint 4 of which are yellow, 2 are red and the remaining jars are brown. Kurt will combine 3 jars of paint into a new container to make a new color which he will name according to the following conditions:

Brun Y if the paint contains 2 jars of brown paint and no yellow.
Brun X if the paint contains 3 jars of brown paint.
Jaune X if the paint contains at least 2 jars of yellow.
Jaune Y if the paint contains exactly 1 jar of yellow.

What is the probability that the new color will be Jaune?

Kurt, a painter, has 9 jars of paint 4 of which are yellow, 2 are red and the remaining jars are brown. Kurt will combine 3 jars of paint into a new container to make a new color which he will name according to the following conditions:

Brun Y if the paint contains 2 jars of brown paint and no yellow. Brun X if the paint contains 3 jars of brown paint. Jaune X if the paint contains at least 2 jars of yellow. Jaune Y if the paint contains exactly 1 jar of yellow.

What is the probability that the new color will be Jaune?

(1) 5/42 (2) 37/42 (3) 1/21 (4) 4/9 (5) 5/9

My answer is C, though I am not good at probability

We can make Jaune color only using option which is marked red, because he takes 3 jars

# of ways to pick any three jars = 9C3 = 84
To get Jaune X --> Either YYX (where X is any color) or YYY
YYX -> 4C2 * 5C1 = 30
YYY -> 4C3 = 4
Total # of ways get Jaune X --> 34

# of ways to get Jaune Y --> YXX (where X is any two colors, apart from yellow)
YXX = 4C1 * 5C2 = 40 ways.

# of ways to pick any three jars = 9C3 = 84 To get Jaune X --> Either YYX (where X is any color) or YYY YYX -> 4C2 * 5C1 = 30 YYY -> 4C3 = 4 Total # of ways get Jaune X --> 34

# of ways to get Jaune Y --> YXX (where X is any two colors, apart from yellow) YXX = 4C1 * 5C2 = 40 ways.

Kurt, a painter, has 9 jars of paint 4 of which are yellow, 2 are red and the remaining jars are brown. Kurt will combine 3 jars of paint into a new container to make a new color which he will name according to the following conditions:

Brun Y if the paint contains 2 jars of brown paint and no yellow. Brun X if the paint contains 3 jars of brown paint. Jaune X if the paint contains at least 2 jars of yellow. Jaune Y if the paint contains exactly 1 jar of yellow.

What is the probability that the new color will be Jaune?

# of ways to pick any three jars = 9C3 = 84 To get Jaune X --> Either YYX (where X is any color) or YYY YYX -> 4C2 * 5C1 = 30 YYY -> 4C3 = 4 Total # of ways get Jaune X --> 34

# of ways to get Jaune Y --> YXX (where X is any two colors, apart from yellow) YXX = 4C1 * 5C2 = 40 ways.

Total # of ways to get Jaune = 74 ways.

P = 74/84 = 37/42

Perfect ! that's the OA (B)

Hi I just did this problem and I'm a bit confused on why my approach is wrong

instead of the following YYX -> 4C2 * 5C1 = 30 YYY -> 4C3 = 4

i just did 4c2 * 7c1 , choosing 2 yellows, and the 3rd being any of the remaining (including yellow)

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