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# M02-06

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Manager
Joined: 26 Feb 2018
Posts: 77
Location: United Arab Emirates
GMAT 1: 710 Q47 V41
GMAT 2: 770 Q49 V47

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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
Manager
Joined: 18 Jul 2018
Posts: 50
Location: United Arab Emirates

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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.

Hi Xylan

I understand logic works perfectly here!

Manager
Status: The darker the night, the nearer the dawn!
Joined: 16 Jun 2018
Posts: 56

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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.

Hi Xylan

I understand logic works perfectly here!

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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Manager
Joined: 18 Jul 2018
Posts: 50
Location: United Arab Emirates

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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.

Hi Xylan

I understand logic works perfectly here!

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 !!!
Re: M02-06   [#permalink] 01 Apr 2019, 06:18

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# M02-06

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