How Many Blue and Red Socks Are in the Drawer?

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In summary, the conversation discusses a problem involving selecting socks from a drawer with a certain probability. The daughter has determined that the probability of selecting a blue sock and then a red sock is 5/21, and the number of blue socks is twice the number of red socks. The conversation ends with the daughter looking for help on how to finish the problem.
  • #1
mathgeek7365
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My daughter is stumped, and so am I.

A drawer contains socks that are either blue or red. There are twice as many blue socks as red socks. If you select two socks from the drawer without replacement, the chance that you choose first a blue sock and then a red sock is 5/21. How many of each color sock was in the drawer before you removed any?

She knows that the probability to first draw a blue socks, then a red sock is 5/21.
The probability to draw a blue sock with the first draw is the number of blue socks
divided by the total number of socks. So, P(B)=b/n
She also knows that the probability to draw a red sock after one blue sock has been
taken out is the number of red socks divided by the total number of socks left after
you take one sock out. So, P(R)=r/(n-1). The probability of drawing a blue and then a red sock is shown as P(B,R)=(b/n)*(r/(n-1))=(5/21).
She also knows that there are twice as many blue socks as red socks, so b=2r.

She can replace b with 2r in the equation. P(B,R)=(2r/n)*(r/(n-1))=(5/21)

Is she correct so far? She would also like to know how to finish the problem. Thanks.
 
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I would say you're on the right track. You do know that $2r=b$. You also, incidentally, know what $n$ is in terms of $r$ and $b$, right?

Your probability is given by
$$\frac{b}{n} \cdot \frac{r}{n-1}=\frac{5}{21}.$$
Now if you substitute everything in, and get it all in terms of one variable, you might be able to solve for one of the variables.
 

FAQ: How Many Blue and Red Socks Are in the Drawer?

What is the "Socks in a drawer problem"?

The "Socks in a drawer problem" is a mathematical probability puzzle, also known as the "Socks and Drawers" or "Socks and Shoes" problem. It involves determining the minimum number of socks that need to be pulled out of a drawer in order to guarantee a matching pair.

What is the origin of the "Socks in a drawer problem"?

The problem was first posed by mathematician William Feller in 1945 as part of his book "An Introduction to Probability Theory and its Applications". It has since become a popular puzzle among mathematicians and students of probability.

What is the solution to the "Socks in a drawer problem"?

The minimum number of socks needed to guarantee a matching pair is three. This solution is based on the pigeonhole principle, which states that if there are n items and m pigeonholes with n > m, then at least one pigeonhole must contain more than one item. In this case, the socks represent the items and the drawers represent the pigeonholes.

Can the "Socks in a drawer problem" be applied to other scenarios?

Yes, the principle behind the "Socks in a drawer problem" can be applied to other scenarios involving probability and combinations. For example, it can be used to determine the minimum number of people needed in a room to guarantee that at least two people share the same birthday.

How does the "Socks in a drawer problem" relate to real life?

The "Socks in a drawer problem" may seem like a purely theoretical puzzle, but it can actually have practical applications. It can be used to illustrate the concept of probability and the importance of considering all possible outcomes in decision making. It also highlights the significance of sample size in statistical analysis.

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