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Probability

Expert replies
by recocollins » Tue Jul 31, 2007 10:29 am
(1) In a drawer of shirts 8 are blue, 6 are green, and 4 are magenta. If Mason draws 2 shirts at random, what is the probabililty at least one of the shirts he draws will be blue?

A. 25/153
B. 28/153
C. 5/17
D. 4/9
E. 12/17

(2) A certain consulting firm employs 8 men and 4 women. In March, 3 employees are selected at random to represent the company at a convention, what is the probability that the representatives will NOT all be men?

A. 14/55
B. 3/8
C. 41/55
D. 2/3
E. 54/55

(3) Kurt, a painter, has 9 jars of paint 4 of which are yellow, 2 are red and the remainng jars are brown. Kurt will combine 3 juars 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 last 2 jars of yellow
Jaune Y if the paint contains exactly 1 jar of yellow

What it the probability that the new color will be Juane?

A. 5/42
B. 37/42
C. 1/21
D. 4/9
E. 5/9


I greatly appreciate the help around these questions. :)
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Source: — Problem Solving |

by RM » Tue Jul 31, 2007 10:58 am
1. In a drawer of shirts 8 are blue, 6 are green, and 4 are magenta. If Mason draws 2 shirts at random, what is the probabililty at least one of the shirts he draws will be blue?

Blue 8; Green 6; Magenta 4; Total 18

Total Number of Ways to draw 2 shirts = 18C2
Number of Ways to draw NO Blue shirts = (6+4)C2 = 10C2

Probability of drawing NO Blue Shirt = 10C2/18C2 = 45/153 = 5/17
Probabilty of drawing atleast 1 Blue Shirt = 1 - 5/17 = 12/17
Choice E

2. A certain consulting firm employs 8 men and 4 women. In March, 3 employees are selected at random to represent the company at a convention, what is the probability that the representatives will NOT all be men?

"Probability that the representatives will NOT all be men" is same as "Probability that the representatives will have atleaset 1 women"

Men 8; Women 4; Total 12

Total Number of ways to select 3 employees from group = 12C3
Number of ways to select ALL men = 8C3

Probablity of selecting all Men = 8C3/12C3 = 14/55
Probability of selecting NOT all Men Or atleast One Women = 1 - 14/55 = 41/55. Choice C.
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by recocollins » Tue Jul 31, 2007 12:00 pm
Thanks a ton. It all makes more sense now.
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by recocollins » Tue Jul 31, 2007 12:10 pm
Thanks a ton. It all makes more sense now.
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by jay2007 » Tue Jul 31, 2007 12:49 pm
For 1 and 2 RM is correct:

For 3:
=====
(3) Kurt, a painter, has 9 jars of paint 4 of which are yellow, 2 are red and the remainng jars are brown. Kurt will combine 3 juars 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 last 2 jars of yellow
Jaune Y if the paint contains exactly 1 jar of yellow

What it the probability that the new color will be Juane?

Juane = J say.

Probability of J = Jx + Jy

Jy (probability of 1 exactly jar of yellow) = ((4c1)*(5c2))/(9c3) = 10/21
Jx = (probability of exactly 2 jars of yellow) + probability of exactly 3 jars of yellow
= ((4c2)*(5c1))/(9c3) + ((4c3)*(5c0))/(9c3)
= 17/42
therefore j = 10/21 + 17/42 = 37/42

Therfore answer is B.
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