+287 My sock drawer contains two red socks, two white socks, and two blue socks. For three days in a row, I pull two socks out of my sock drawer at random. (I do not replace the socks in the drawer before drawing on the next day.) What is the probability that for each of the three days my socks are unmatched?
+506 The probability of drawing unmatched socks on any given day depends on the number of socks remaining and the number of matching pairs left. Let's analyze this step-by-step:
Day 1:
Total socks: 6 (2 red, 2 white, 2 blue)
Favorable outcomes (unmatched socks): You can draw any one sock and then a sock from a different pair. There are 4 choices for the first sock (any of the 6 colors) and then 4 remaining socks (excluding the one already drawn). So, there are 4 * 4 = 16 favorable outcomes.
Total possible outcomes: You can draw any two socks from the 6 available. There are 6 choices for the first sock and then 5 remaining socks, resulting in 6 * 5 = 30 total possible outcomes.
Probability on Day 1:
(Favorable outcomes on Day 1) / (Total possible outcomes on Day 1) = 16 / 30 = 8/15
Day 2:
Total socks remaining: 4 (after drawing 2 on Day 1, not replaced)
Favorable outcomes (unmatched socks): Similar to Day 1, you can draw any sock and then one from a different remaining pair. There are 3 choices for the first sock and then 2 remaining socks (excluding the one drawn), resulting in 3 * 2 = 6 favorable outcomes.
Total possible outcomes: There are 4 choices for the first sock and then 3 remaining, resulting in 4 * 3 = 12 total possible outcomes.
Probability on Day 2 (given unmatched socks on Day 1):
We only consider the scenario where you drew unmatched socks on Day 1 because the prompt asks for the probability of this happening for all three days.
So, we only consider the drawers where we have 4 socks remaining (unmatched).
(Favorable outcomes on Day 2) / (Total possible outcomes on Day 2) = 6 / 12 = 1/2
Day 3:
Total socks remaining: 2 (after drawing 2 on Day 2, not replaced)
Favorable outcomes (unmatched socks): There's only one way to draw unmatched socks at this point - you must draw the two remaining socks, which are inherently unmatched.
Total possible outcomes: There are 2 choices for the first sock and then 1 remaining, resulting in 2 * 1 = 2 total possible outcomes.
Probability on Day 3 (given unmatched socks on Days 1 & 2):
Similar to Day 2, we only consider the scenario where you drew unmatched socks on both previous days.
(Favorable outcomes on Day 3) / (Total possible outcomes on Day 3) = 1 / 2
Overall Probability:
The prompt asks for the probability of getting unmatched socks for all three days. To get this probability, we need to multiply the probabilities of getting unmatched socks on each day (considering the condition that you drew unmatched socks on the previous day).
Overall Probability = (Probability on Day 1) * (Probability on Day 2 | Day 1) * (Probability on Day 3 | Day 1 & 2) = 8/15 * 1/2 * 1/2 = 8 / 60
Simplifying the fraction:
We can divide both the numerator and denominator by 4:
Overall Probability = 2 / 15
Therefore, the probability of drawing unmatched socks for all three days is 2/15.
+42 I just wanted to note that booboo44's solution may be incorrect. First off, there are 6 choose 2 = 15 possible outcomes, not 30. This is because cases will be counted twice. In booboo44's solution, AB and BA are treated as different cases, when they are two equal cases. The favorable cases should also be 12, not 16. There may be more mistakes later on in the solution (but I won't be covering them). If all calculations are done right, the answer should be 8/15. (I created a solution but apparently it's still being moderated)
This is for educational purposes only. I do not want to start a fight with anyone.
Feel free to tell me if I did anything wrong! :D
