You have a drawer with $6$ loose blue socks, and $10$ loose brown socks. If you grab two socks from the drawer in the dark (random draw), what is the probability that you draw a brown pair?
I have $\frac{5}{8}=\frac{10}{16}=.625$.
[Math] Probability of drawing a pair of brown socks
probability
Related Solutions
Option A:
- The total number of ways to select $3$ socks is $\binom{4+6+8}{3}=816$
- The number of ways to select $1$ sock of each color is $\binom41\cdot\binom61\cdot\binom81=192$
- Hence the probability for at least $2$ socks of the same color is $1-\frac{192}{816}$
Option B:
- The total number of ways to select $3$ socks is $\binom{4+6+8}{3}=816$
- The number of ways to select $3$ brown is $\binom43 = 4$
- The number of ways to select $3$ blue is $\binom63 = 20$
- The number of ways to select $3$ yellow is $\binom83 = 56$
- The number of ways to select $2$ brown and $1$ blue is $\binom42\cdot\binom61= 36$
- The number of ways to select $2$ brown and $1$ yellow is $\binom42\cdot\binom81= 48$
- The number of ways to select $2$ blue and $1$ brown is $\binom62\cdot\binom41= 60$
- The number of ways to select $2$ blue and $1$ yellow is $\binom62\cdot\binom81=120$
- The number of ways to select $2$ yellow and $1$ brown is $\binom82\cdot\binom41=112$
- The number of ways to select $2$ yellow and $1$ blue is $\binom82\cdot\binom61=168$
- This adds up to $624$ ways for selecting at least $2$ socks of the same color
- Hence the probability for at least $2$ socks of the same color is $\frac{624}{816}$
There are three cases to consider:
- Four socks, each of a different color, are selected on Monday and Tuesday, then a pair is selected on Wednesday from the one pair and four single socks that remain. The probability of this occurring is $$1 \cdot \frac{8}{9} \cdot \frac{6}{8} \cdot \frac{4}{7} \cdot \frac{\binom{2}{2}}{\binom{6}{2}}$$ since the first sock is guaranteed to be of a distinct color, $8$ of the remaining $9$ socks are a different color than that of the first sock that is selected, $6$ of the remaining $8$ socks are of a different color than those of both of the first two socks that have been selected, $4$ of the remaining $7$ socks are of a different color than those of the first three socks that have been selected, and we must choose both socks of the fifth color from the six remaining socks.
- Four socks, of three different colors, are selected on Monday and Tuesday, then a pair is selected on Wednesday from the two remaining pairs and two remaining individual socks. The probability of this occurring is $$1 \cdot \frac{8}{9} \cdot \left(\frac{2}{8} \cdot \frac{6}{7} + \frac{6}{8} \cdot \frac{2}{7}\right) \cdot \frac{4}{6} \cdot {1}{5}$$ On Monday, the probabilities are the same as in the first case. On Tuesday, there are two possibilities. One is that a sock of one of the $2$ colors already selected on Monday is selected from one of the $8$ remaining socks, then a sock of one of the other $3$ colors is selected from the $7$ remaining socks. The other possibility is that one of the $6$ socks of one the $3$ remaining colors is selected from the $8$ socks that remain, then one of the $2$ socks of a color already selected on Monday is selected from one of the $7$ remaining socks. On Wednesday, there are $6$ socks left. One of the $4$ socks from the two remaining pairs must be selected first, then the only sock of that color that remains must be selected from the $5$ remaining socks.
- Four socks, of two different colors, are selected on Monday and Tuesday, then a pair is selected on Wednesday from one of the three remaining pairs. The probability of this occurring is $$1 \cdot \frac{8}{9} \cdot \frac{2}{8} \cdot \frac{1}{7} \cdot 1 \cdot \frac{1}{5}$$ On Monday, the probabilities are the same as in the preceding cases. On Tuesday, the first sock selected must be one of the $2$ socks of a color already selected on Monday, then the other sock that is selected must match the other color selected on Monday. On Wednesday, only the three colors that have not already been used are available, so the first sock is guaranteed to be of a different color than those selected on Monday and Tuesday. Finally, we must select the only remaining sock that matches the first selection on Wednesday from the five remaining socks.
Since these cases are mutually exclusive, you can add the probabilities to determine the probability that Wednesday is the first day that Sandy selects a matching pair of socks.
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Best Answer
The easier solution (for understanding) would be:
$$P=\frac{C_2^{10}}{C_2^{16}}=\frac{\frac{10!}{2!(10-2)!}}{\frac{16!}{2!(16-2)!}}=0.375$$
That is, what you want is to pick 2 of those 10 brown socks, and it is possible that you choose any 2 of all 16 socks.