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Before leaving for his business trip, Chad asks his assistant to choos

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Before leaving for his business trip, Chad asks his assistant to choos  [#permalink]

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New post 02 Dec 2014, 21:50
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  55% (hard)

Question Stats:

65% (02:11) correct 35% (02:43) wrong based on 213 sessions

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Before leaving for his business trip, Chad asks his assistant to choose and pack four shirts from his closet, which currently contains eight shirts. If each shirt is a different color, including one blue shirt and one pink shirt, and the assistant chooses the shirts at random, what is the probability that the pink shirt will be one of the four packed but the blue shirt will not?

A 4/7
B 1/2
C 27/70
D 2/7
E 9/35
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Re: Before leaving for his business trip, Chad asks his assistant to choos  [#permalink]

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New post 17 Jun 2015, 21:09
6
4
anceer wrote:
Before leaving for his business trip, Chad asks his assistant to choose and pack four shirts from his closet, which currently contains eight shirts. If each shirt is a different color, including one blue shirt and one pink shirt, and the assistant chooses the shirts at random, what is the probability that the pink shirt will be one of the four packed but the blue shirt will not?

A 4/7
B 1/2
C 27/70
D 2/7
E 9/35



Responding to a pm:

Probability = Favorable Cases/Total Cases

Favorable Cases: You need 4 shirts. Choose pink shirt. You have already got one. You need three more. Put the blue shirt away since it is not an option.
So you have 6 possible options (8 - pink - blue) for the shirts and you need to choose 3 of them (to get a total of 4 shirts). You can choose 3 shirts out of 6 in 6C3 ways.
6C3 = 6*5*4/3*2 = 20 ways
There are 20 ways of choosing a pink shirt and three others from the remaining (except blue).

Total Cases: Choose any 4 shirts out of 8 in 8C4 ways.
8C4 = 8*7*6*5/4*3*2 = 70

Probability = 20/70 = 2/7
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Re: Before leaving for his business trip, Chad asks his assistant to choos  [#permalink]

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New post 03 Dec 2014, 01:21
1
1
anceer wrote:
Before leaving for his business trip, Chad asks his assistant to choose and pack four shirts from his closet, which currently contains eight shirts. If each shirt is a different color, including one blue shirt and one pink shirt, and the assistant chooses the shirts at random, what is the probability that the pink shirt will be one of the four packed but the blue shirt will not?

A 4/7
B 1/2
C 27/70
D 2/7
E 9/35


out of 8 shirts, any four shirts can be chosen in 8C4 ways.
as per the question pink shirt should be among four selected shirts. thus number of ways selecting a blue shirt=1
also, we don't want blue shirt among any of the four selected shirts. thus the remaining 3 shirts will be selected from 6{why?. because one pink shirt is already selected and we cannot have blue shirt among the selected shirts so 8-2=6) available shirts. this can be done in 6C3 ways.

thus required probability = \(\frac{(1.6C3)}{(8C4)}\)
= \(\frac{2}{7}\)
hence D
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Re: Before leaving for his business trip, Chad asks his assistant to choos  [#permalink]

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New post 08 Dec 2014, 21:02
required prob (1 pink + other colors except blue) = (no. of ways to get 1 pink + other colors except blue) / (total ways of choosing 4 out of 8 shirts )

(no. of ways to get 1 pink + other colors except blue) --> using slots method: we have 4 slots
1(1st slot for pink shirt) 6(as out of 8 shirts: 1 pink is already selected and we cannot select blue) 5 4 = 5

we can get this in 4 different ways, hence 5*4 = 20 ...........numerator

We can get any 4 shirts out of 8 (without any restrictions) in 70 ways ..................denominator

required prob (1 pink + other colors except blue) = (no. of ways to get 1 pink + other colors except blue) / (total ways of choosing 4 out of 8 shirts )
= 20/70
= 2/7

Ans. D) 2/7
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Re: Before leaving for his business trip, Chad asks his assistant to choos  [#permalink]

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New post 20 Jun 2015, 12:38
Could anyone suggest what is wrong in my calculations:

(1*6*5*4)/(8*7*6*5)=1/14
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Before leaving for his business trip, Chad asks his assistant to choos  [#permalink]

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New post 20 Jun 2015, 13:27
2
Boycot wrote:
Could anyone suggest what is wrong in my calculations:

(1*6*5*4)/(8*7*6*5)=1/14


You have the right answer to a slightly different question - if a person picks four shirts one at a time, what is the probability he or she picks the pink shirt first, and then picks three non-blue shirts?

That is, in your numerator, you've counted the number of ways of picking a Pink shirt with your first selection, then three shirts of a non-Blue colour with the remaining three selections. But in this question we don't need to pick the Pink shirt first - we might instead pick it second, third or fourth. So you need to multiply your numerator by 4, and then you'll have the right answer.
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Re: Before leaving for his business trip, Chad asks his assistant to choos  [#permalink]

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New post 21 Jun 2015, 01:17
Thanks Ian.

Yes, I forgot to multiply by 4!/3!
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Re: Before leaving for his business trip, Chad asks his assistant to choos  [#permalink]

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New post 23 Jul 2018, 17:19
anceer wrote:
Before leaving for his business trip, Chad asks his assistant to choose and pack four shirts from his closet, which currently contains eight shirts. If each shirt is a different color, including one blue shirt and one pink shirt, and the assistant chooses the shirts at random, what is the probability that the pink shirt will be one of the four packed but the blue shirt will not?

A 4/7
B 1/2
C 27/70
D 2/7
E 9/35


The number of ways to select the pink shirt but not blue shirt is 1C1 x 1C0 x 6C3 = (6 x 5 x 4)/3! = 20. (Notice that 1C1 is for the pink shirt to be selected, 1C0 is for the blue shirt not to be selected, and 6C3 is the number of ways to select the 3 shirts from the remaining 6 shirts.)

The number of ways to select 4 shirts from 8 shirts is:
8C4 = 8!/(4! x 4!) = (8 x 7 x 6 x 5)/4! = (8 x 7 x 6 x 5)/(4 x 3 x 2) = 7 x 2 x 5 = 70

Thus, the probability that the pink shirt will be one of the four packed but the blue shirt will not is 20/70 = 2/7.

Answer: D
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Re: Before leaving for his business trip, Chad asks his assistant to choos &nbs [#permalink] 23 Jul 2018, 17:19
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