In a drawer of shirts 8 are blue, 6 are green and 4 are

Question

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 probability 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

upamanyu · Accepted Answer

I think its E

there are a total of 18 shirts. 8 blue and 10 non blue.

P(selecting at least 1 blue shirt) = 1 - P(selecting no blue shirts)

Assuming no replacement

P(selecting first non-blue shirt) = 10/18
P(selecting secong non-blue shirt) = 9/17

P(selecting no blue shirts) = 10/18 * 9/17= 10/34

Therefor P(selecting at least 1 blue shirt) = 1 -(10/34)= 24/34=12/17

TeHCM · Answer

Does it matter if he draws the shirts at once or in order.....

krisrini · Answer

= 1 - p(None of them being blue)
= 1 - (10/18 * 9 /17)
= 1 - 5/17
= 12/17 E

Bunuel · Answer

Mathematically the probability of picking 2 shirts simultaneously, or picking them one at a time (without replacement) is the same.

kona · Answer

Hi Bunuel,
Please find my doubt below.

q:In a drawer of shirts, 8 are blue, 6 are green, and 4 are red. If Miles draws two shirts at random, what is the probability that at least one of the shirts he draws is blue?

My approach:1-(zero shirts blue)

1-(10*9)/18c2

1-(10/17)=7/17.

Wrong..

Another approach:1-(10/18*9/17)

1-(5/17)=12/17

Right..

My doubts:1.When picking two shirts at random from a group of 18,is it not 18c2 for a total number of possibilities

2.Also,does "picking at random" means "picking simultaneously?"

3.Does ordering in this case matter?When should the order be considered and when it shouldnt be?Could you please throw some light on this?

Please rectify

Thank you

Regards,
Kona

Bunuel · Answer

1. If you use combination approach then it should be P(at \ least \ one \ blue) = 1- P(0 \ blue) = 1 - \frac{C^2_{10}}{C^2_{18}} = \frac{12}{17} . So, your denominator was correct but the numerator was not. It should have been C^2_{10} : the number of ways to pick two not-blue shirts out of 10. 2. No. You can pick randomly one-by-one as well as simultaneously. 3. Theory on Combinations: math-combinatorics-87345.html DS questions on Combinations: search.php?search_id=tag&tag_id=31 PS questions on Combinations: search.php?search_id=tag&tag_id=52 Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html Theory on probability problems: math-probability-87244.html All DS probability problems to practice: search.php?search_id=tag&tag_id=33 All PS probability problems to practice: search.php?search_id=tag&tag_id=54 Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html Hope this helps.

ScottTargetTestPrep · Answer

We can use the equation:

P(at least one of the shirts he draws will be blue) = 1 - P(no blue shirts)

P(no blue shirts) = 10/18 x 9/17 = 5/9 x 9/17 = 5/17.

So, P(at least one of the shirts he draws will be blue) = 1 - 5/17 = 12/17.

Answer: E

kop18 · Answer

I assumed this question to be similar with the usual "picking 5 green balls and 3 blue balls" sort of questions in which every ball is the same and it does not matter which ball u are picking. Usually in these kind of questions unless we are told explicitly that the balls have been picked simultaneously or that repetition is not allowed - we must assume that repetition is allowed.

Logically, when I look at this question it makes sense that once he's picked a shirt he will not put that same shirt back and then make the second pick. So if someone could please explain how I can ensure that the above doubt does not prop in my head when I am looking at these kinds of question then that'd be helpful -  Bunuel   KarishmaB

Bunuel · Answer

A proper GMAT question would make it clear whether selection is with or without replacement.

kop18 · Answer

Thank you for confriming this

I broke my head for too long trying to understand why the solutions above just assumed probablity of picking shirt w/o replacemaent. This was helpful

luisdicampo · Answer

Deconstructing the Question

Topic: Probability - "At Least One" Shortcut

Mason draws 2 shirts from a drawer of  18 total shirts  (8 Blue, 6 Green, 4 Magenta). We need the probability that  at least one  is Blue.

1. Identify the Complement  
The complement of "at least one Blue" is  "Zero Blue shirts"  (meaning both shirts selected are from the Green or Magenta pool).
 
 Total shirts = 18 
 Non-Blue shirts = 6 (Green) + 4 (Magenta) = 10

2. Calculate P(No Blue)  
The probability of picking two non-blue shirts in a row:
 P(None) = (10/18) * (9/17) 
 P(None) = (5/9) * (9/17) =    5/17

3. Final Calculation  
 P(At least one Blue) = 1 - P(None) 
 P = 1 - 5/17 =    12/17

The correct answer is E.

In a drawer of shirts 8 are blue, 6 are green and 4 are

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