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

Question

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

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

KarishmaB · Accepted Answer

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

manpreetsingh86 · Answer

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

JJo · Answer

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

Boycot · Answer

Could anyone suggest what is wrong in my calculations:

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

IanStewart · Answer

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.

Boycot · Answer

Thanks Ian.

Yes, I forgot to multiply by 4!/3!

ScottTargetTestPrep · Answer

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

Bambi2021 · Answer

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

grindingtedd · Answer

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

psakshi896 · Answer

(30*4)/(56*30)= 4/56= 2/18= 1/9

Hubble · Answer

Hey Ian, Since all 4 shirts are different shouldn't we be multiplying by 4!. I'm unable to understand the logic for dividing by 3! as well.

Bunuel · Answer

We multiply by 4!/3! = 4 because there are 4 possible positions for the pink shirt to be selected.

Here's the key part: The sequence of pink, non-blue, non-blue, non-blue can be arranged in 4!/3! = 4 different ways since the pink shirt can be in the 1st, 2nd, 3rd, or 4th position.

You don't multiply by 4! (which would be for arranging all 4 shirts in different ways) because the order within the non-blue shirts doesn’t matter. You are only concerned about where the pink shirt is placed.

So, multiplying by 4 accounts for the 4 positions the pink shirt can occupy.

Jalmagro · Answer

Find the total scenarios first.

8C4 = 8!/4!4! = 70

Then find the total choices of Pink being selected

7C3 = 7!/4!3! = 35

Then find the total choices of both being selected together

6C2 = 6!/4!2! = 15

Subtract the total choices for both from the total choices for Pink

35 - 15 = 20

Place the total for only Pink over the total number of choices to get the probability

20/70, simplify, 2/7

Answer: D:)

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

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