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Re: Before leaving for his business trip, Chad asks his assistant to choos [#permalink]
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]
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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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]
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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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 [#permalink]
I had a different approach:

4/8 to get the pink shirt

remaining is 4/7 spots to not get the blue shirt

1/2x4/7 = 4/14 = 2/7
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Re: Before leaving for his business trip, Chad asks his assistant to choos [#permalink]
It would be the Probability of choosing the pink shirt minus the probability of choosing both the pink and blue shirt

the probability of choosing 3 shirts randomly out of 7 shirts (since pink is taken) minus the probability of choosing 2 shirts randomly out of 6 shirts (since pink and blue are taken)


(7C3-6C2)/8C4
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Re: Before leaving for his business trip, Chad asks his assistant to choos [#permalink]
Boycot wrote:
Could anyone suggest what is wrong in my calculations:

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



(30*4)/(56*30)= 4/56= 2/18= 1/9
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Re: Before leaving for his business trip, Chad asks his assistant to choos [#permalink]
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