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A statistician keeps a simple wardrobe. He only purchases pairs of black socks and white socks, and he keeps all of the socks in a pile in the drawer. Recently one of the socks was lost in the laundry. The socks have a mathematical property. If you select two socks at random from the drawer, the socks will match in color exactly 50% of the time. The statistician owns more than 200 socks but less than 250 socks, and there are more black socks than white socks. How many socks of each color are there? And which color sock was lost in the laundry? Thanks for help.

Guest Jun 24, 2017

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 #1
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Let the number of black socks=b
 Let the number of white socks=w
On the first draw, we can draw either a black sock or a white sock. On the next draw, we could draw a black sock, in wich case will have: b - 1 black socks remaining. Then: b(b - 1) = Total number of ways we could have drawn 2 black socks in a row.
On the second draw, we could have drawn a white sock, in which case we would have w white socks remaining. And if we multiplied bw = number of ways we could have drawn a black on the first draw and a white on the second draw.
Now, let us suppose on the first draw we drew a white sock. On the second draw, we could draw either a white or a black sock. If it is a black sock, then we have bw=number of ways we could have drawn a white sock first followed by a black sock.
But we could have also drawn a white sock on the second draw. Since we had already drawn a white sock on the first draw, then we have: w-1 socks remaining. And w(w-1) =number of ways we could have drawn 2 white socks in a row.
Now we turn to number crunching!!
We know the probability of picking 2 matching socks =50%, which means the probability of NOT picking 2 matching socks=50%. In other words, the number of ways of picking a matching pair = number of ways of  NOT picking a matching pair.
Now we have already determined the number of ways of picking a matching pair of both colors, namely: b(b-1) + w(w-1). The remaining possibility is the picking of 2 socks which do NOT match, and that is =bw + bw =2bw. So now we have this equality:
b(b-1) + w(w-1) =2bw. By simple algebraic manipulation we get: (b-w)^2 = b+w. Now, this equation gives us a clue that the number of socks(b+w) is a perfect square! But we are also told that the number of socks is:200<(b+w)<250. Now we can consider which number is a perfect square that is between 200 and 250? Well: 14^2 =196, 15^2 =225 and 16^2=256. So 15^2 = 225, appears to be the number we looking for. So the total number of socks (b+w) =225. We also know that:(b-w)^2 =15^2. This implies that either b=15 or w=15. But we are told that there are MORE black socks than white socks, or b-w = 15. So now we have a system of equations: (b+w) =225 and b-w = 15. Solving for b and w, we get b=120 and w=105.
Finally, the missing sock!!. Well, we know that the "sloppy statistician" buys socks in "pairs!!". Therefore, the missing sock must be white, since there are only 105 of them, which an ODD number!.
And that is the end of this beautiful problem. Congrats to the questioner who came up with this wonderful question! I racked my brains for an hour or so. I hope I got it right!!.

Guest Jun 24, 2017
edited by Guest  Jun 24, 2017
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4+0 Answers

 #1
avatar
+1
Best Answer

Let the number of black socks=b
 Let the number of white socks=w
On the first draw, we can draw either a black sock or a white sock. On the next draw, we could draw a black sock, in wich case will have: b - 1 black socks remaining. Then: b(b - 1) = Total number of ways we could have drawn 2 black socks in a row.
On the second draw, we could have drawn a white sock, in which case we would have w white socks remaining. And if we multiplied bw = number of ways we could have drawn a black on the first draw and a white on the second draw.
Now, let us suppose on the first draw we drew a white sock. On the second draw, we could draw either a white or a black sock. If it is a black sock, then we have bw=number of ways we could have drawn a white sock first followed by a black sock.
But we could have also drawn a white sock on the second draw. Since we had already drawn a white sock on the first draw, then we have: w-1 socks remaining. And w(w-1) =number of ways we could have drawn 2 white socks in a row.
Now we turn to number crunching!!
We know the probability of picking 2 matching socks =50%, which means the probability of NOT picking 2 matching socks=50%. In other words, the number of ways of picking a matching pair = number of ways of  NOT picking a matching pair.
Now we have already determined the number of ways of picking a matching pair of both colors, namely: b(b-1) + w(w-1). The remaining possibility is the picking of 2 socks which do NOT match, and that is =bw + bw =2bw. So now we have this equality:
b(b-1) + w(w-1) =2bw. By simple algebraic manipulation we get: (b-w)^2 = b+w. Now, this equation gives us a clue that the number of socks(b+w) is a perfect square! But we are also told that the number of socks is:200<(b+w)<250. Now we can consider which number is a perfect square that is between 200 and 250? Well: 14^2 =196, 15^2 =225 and 16^2=256. So 15^2 = 225, appears to be the number we looking for. So the total number of socks (b+w) =225. We also know that:(b-w)^2 =15^2. This implies that either b=15 or w=15. But we are told that there are MORE black socks than white socks, or b-w = 15. So now we have a system of equations: (b+w) =225 and b-w = 15. Solving for b and w, we get b=120 and w=105.
Finally, the missing sock!!. Well, we know that the "sloppy statistician" buys socks in "pairs!!". Therefore, the missing sock must be white, since there are only 105 of them, which an ODD number!.
And that is the end of this beautiful problem. Congrats to the questioner who came up with this wonderful question! I racked my brains for an hour or so. I hope I got it right!!.

Guest Jun 24, 2017
edited by Guest  Jun 24, 2017
 #2
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Wow!!!.....very impressive, guest..........I followed your logic and it seems reasonable.....!!!

 

 

 

cool cool cool

CPhill  Jun 24, 2017
 #3
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Thank you CPhill. You are very kind. By the way, there may be shortcuts to this, but I still haven't arrived at one yet!.

Guest Jun 24, 2017
 #4
avatar+76837 
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Eh....I have a feeling that your answer was about as good as it gets.....very well laid out and explained....!!!

 

 

 

cool cool cool

CPhill  Jun 24, 2017

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