Question 1060190: 1) I have the following individual socks that are in my drawers: 6 red socks, 7 green socks, and 4 blue socks. Two socks are picked are picked out in the dark.
a) what is the probability that if the second sock is red, then the first sock is also red? (note: we can also phrase this as "what is the probability that the first sock was red given the seond sock is red.)
b) what is the probability that the two socks are the same color?
c) what is the probability that if the second socks are not red, then the first sock is blue? (this can also be phrased as "what is the probability that the first sock was blue given that the second sock is not red(is blue or green).")
Found 2 solutions by Boreal, KMST:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Given that the first sock was red, there are 5 red, 7 green, and 4 blue socks left.
The probability the second sock is red is 5/16.
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probability both are red is 6/17*(5/16)=30/272
probability both are green is (7/17)(6/16)=42/272
probability both are blue is (4/17)*3/16)=12/272
That sum is 84/272, or 21/68.
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the possibilities are blue/blue, blue/green, green/blue, green/green, red/blue, red/green. All denominators are 272
B-B:12
B-G:28
G-B:28
G-G:42
R-B:24
R-G:42
182
182/272.
Of those 182, 40 are blue. The probability is 40/182=0.220
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a rough way to check it is that the probability the first sock is blue is 4/17 or 0.235. If we look at what the second sock can't be red, then being blue is one possibility. If that is the case, it lessens the likelihood of the first sock's also being blue.
Answer by KMST(5328)  (Show Source):
You can put this solution on YOUR website! It is easy to get mixed up and make a mistake in the cslculations, so I made a diagram.
All the possibilities can be represented in the diagram below.
The numbers next to the left edge represent the number of ways to pick the first sock in each color.
If the first sock was put back before picking a second sock,
The diagram would be a perfect square, divided into even rows and columns.
Since you do not replace, the diagram is a rectangle with slightly mismatched columns.
The numbers along the horizontal lines show the width of each column.
The number of ways to get any color sequence for the socks picked
is the "area" of the corresponding rectangle
(the product of its width and height numbers).
On paper, I wrote the "area" inside each of the small rectangles,
and I also added the areas of each row and column.
(The numbers are repeated).
I did not write that much in the diagram, because it takes a lot of time to doing it through the website.
For example, there are 4 ways to get a blue sock in the first draw,
and after that there are 3 blue socks left.
That is shown as the height and width of the middle rectangle.
Their product,  is the number of ways to pick 2 blue socks.
You could count them if you had a way to tell one blue sock apart from another.
Similarly,the number of ways to pick a red sock first, and then a blue one is
 , which is represented by the 6 by 4 rectangle in the middle of the top row.
The case of a red second sock is represented by the (uneven) left column.
The top left rectangle represents the  ways to get two red socks.
The rectangle below that shows the  ways to get a blue sock first, and then a red one.
The rectangle beow that represents the  ways to get a green sock first, and then a red one.
The total ways of ending with a red sock for the second draw is
 , and only in  of those cases were both socks red.
That means, the answer to part a) is  .
For part c), the whole big rectangle represents the  ways to get two socks in any color sequence.
Since there were  ways to have get a red sock in the second draw,
there are  ways to get a second sock that is not red,
represented by the two left columns.
The part of the blue middle row in the two left (non-red) columns represents
the  or  ways to get a first blue sock,
without getting a second red sock.
Those  ways out of the  ways of avoiding a second red sock are
 .
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