SOLUTION: There are three different kinds of socks in a bag: red, blue, and yellow. The probability of choosing a red sock is 1/5. If there are twice as many blue socks as red socks, what is

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Question 260011: There are three different kinds of socks in a bag: red, blue, and yellow. The probability of choosing a red sock is 1/5. If there are twice as many blue socks as red socks, what is the probability of choosing a yellow sock?
Found 2 solutions by edjones, Edwin McCravy:
Answer by edjones(8007)   (Show Source): You can put this solution on YOUR website!
Assume that there is only one red sock. Then there are 2 blue socks. Then there must be 2 yellow socks so that the probability of choosing a red sock is 1/5. So the probability of choosing a yellow sock is 2/5.
.
Ed

Answer by Edwin McCravy(20062)   (Show Source): You can put this solution on YOUR website!
There are three different kinds of socks in a bag: red, blue, and yellow. The probability of choosing a red sock is 1/5. If there are twice as many blue socks as red socks, what is the probability of choosing a yellow sock?


th of them are red, ths or them are blue.

That means ths of them are either red or blue, so

the remaining  have to be yellow, so the probability
of getting a yellow sock is ths.
----------------------------------------------
Or you could look at it this way:   

Suppose there are R red socks, B blue socks and Y yellow socks.

Then there are R+B+Y socks in the bag.

Since the probability of getting a red sock is ,



Cross-multiplying:





Since there are twice as many blue socks as red socks, 

so if we substitute that:





So there are twice as many yellow socks as red socks,
so you're twice as likely to select a yellow one as a
red one, so twice  is .

Answer: 

Checking: There could be just 5 socks in the bag, 1 red ones,
2 blue ones and 2 yellow ones.  There are twice as many blue
ones than red ones, and the probability of getting a red one
is , and the probability of getting a blue one is the
sam as that of getting a yellow one, which is is {{2/5}}}.

Edwin

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