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M02-06

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Re: M02-06  [#permalink]

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New post 10 Mar 2018, 15:05
Method for this question


When ever you see "at least" this is a hint for using the formula P (at least) = 1 - (P less than) (which is easier than considering all the individual scenarios for 2, 3 or 4 matching socks and summing them. That would be very time consuming as we could have WWWW, WWWB, WWWG, WWBB, WWGG WWBG BBWG BBGG BBBW BBBG GGBW etc)

So what is the probability of NOT picking at least 2 socks of the same colour? IE less than 2 matching socks? Ie. NO matching socks. For that we need all 4 socks of a different colour. However, we only have 3 colours, so the probability of this is zero. 1-0=1
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Re: M02-06  [#permalink]

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New post 31 Mar 2019, 01:27
Bunuel wrote:
Official Solution:

There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four individual socks are picked at random what is the probability of getting at least two socks of the same color?

A. \(\frac{1}{5}\)
B. \(\frac{2}{5}\)
C. \(1\)
D. \(\frac{3}{5}\)
E. \(\frac{4}{5}\)


No formula is need to answer this question. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.


Answer: C



Hi Xylan

I understand logic works perfectly here!

What about complement rule??


THANKS IN ADVANCE
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M02-06  [#permalink]

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New post 31 Mar 2019, 07:50
2
JIAA wrote:
Bunuel wrote:
Official Solution:

There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four individual socks are picked at random what is the probability of getting at least two socks of the same color?

A. \(\frac{1}{5}\)
B. \(\frac{2}{5}\)
C. \(1\)
D. \(\frac{3}{5}\)
E. \(\frac{4}{5}\)


No formula is need to answer this question. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case, we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.


Answer: C



Hi Xylan

I understand logic works perfectly here!

What about complement rule??


THANKS IN ADVANCE


Given Data:
    10 White | 6 Black | 4 Grey

JIAA By the complement rule:
Total = Case-1 (when probability is at least 2) + Case-2 (When probability is less than 2)

To purposefully find the probability for which we get less than 2 socks of the same color, Let's fill the 3 of the 4 empty spaces:
    W - B - G - __ 4th-space __

As we can see, the 4th-space can ONLY be filled by any of the already existing colors of the socks , i.e., by White, Black OR Grey, thus NO scenario is possible that can fit the conditon.
    Hence, the probability for which we get less than 2 socks of the same color = \(0\) - The Complementing Case

Therefore, ALL the possible cases have at least 2 socks of the same colors, i.e., Probability = 1
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Re: M02-06  [#permalink]

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New post 01 Apr 2019, 06:18
Xylan wrote:
JIAA wrote:
Bunuel wrote:
Official Solution:

There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four individual socks are picked at random what is the probability of getting at least two socks of the same color?

A. \(\frac{1}{5}\)
B. \(\frac{2}{5}\)
C. \(1\)
D. \(\frac{3}{5}\)
E. \(\frac{4}{5}\)


No formula is need to answer this question. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case, we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.


Answer: C



Hi Xylan

I understand logic works perfectly here!

What about complement rule??


THANKS IN ADVANCE


Given Data:
    10 White | 6 Black | 4 Grey

JIAA By the complement rule:
Total = Case-1 (when probability is at least 2) + Case-2 (When probability is less than 2)

To purposefully find the probability for which we get less than 2 socks of the same color, Let's fill the 3 of the 4 empty spaces:
    W - B - G - __ 4th-space __

As we can see, the 4th-space can ONLY be filled by any of the already existing colors of the socks , i.e., by White, Black OR Grey, thus NO scenario is possible that can fit the conditon.
    Hence, the probability for which we get less than 2 socks of the same color = \(0\) - The Complementing Case

Therefore, ALL the possible cases have at least 2 socks of the same colors, i.e., Probability = 1



THANK YOU Xylan

MUCH APPRECIATED !!!
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Re: M02-06  [#permalink]

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New post 11 Jun 2019, 03:00
I got this wrong, but I didn't think it was this simple when reviewing the explanation.

I think a lot of people are over complicating this.

There are 10 White, 6 Blue and 4 Grey.

What is the probability that I pick a matching pair of socks if I am to choose 4 individual socks?
4 Choices: _ _ _ _

Randomize this: W B G ?
? = we must choose either White, Blue or Grey. So we have a 100% chance of picking a matching pair we choose
W B G (W/B/G) = Matching pair.
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Re: M02-06  [#permalink]

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New post 11 Jun 2019, 05:58
chetan2u wrote:
MEOWSER wrote:
Bunuel,

Thanks for the question. If it were phrased differently and we only picked 3 individual socks instead of 4 individual socks, how would we approach the question?


the recommended way would be..
in this case you will have to find the probability of all three being different...
since there are three different socks , the ways to pick up three amongst th three would be 3!=6...
now the probability in each way=\(\frac{10}{20}*\frac{6}{19}*\frac{4}{18}\)=\(\frac{2}{57}\)...
so for 6 ways=6*\(\frac{2}{57}\)= \(\frac{12}{57}\)..
so the prob of not having all different or in other words having atleast two of same colour= 1-\(\frac{12}{57}\)=\(\frac{45}{57}\)


Hello
Please explain how does the selection order (and thereby an extra factor of 3!) matter in the above solution provided if we were to make a selection of 3 individual socks.

Thanks
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Re: M02-06   [#permalink] 11 Jun 2019, 05:58

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