hydr01 wrote:
Bunuel wrote:
A drawer contains 5 pairs of white socks, 3 pairs of black socks, and 2 pairs of grey socks. If four individual socks are randomly chosen without replacement, what is the probability of getting at least two socks of the same color?
A. \(\frac{1}{5}\)
B. \(\frac{2}{5}\)
C. \(1\)
D. \(\frac{3}{5}\)
E. \(\frac{4}{5}\)
If we were to use a formula, what formula would we use? Would the solution be a mix of combinations and probability formulas? Can you please walk through what using a formula would look like? If it's easier, what formula would we use if there there a fourth color pair of socks in the mix?
WWWWWWWWWW - BBBBBB - GGGG
We need to find the probability of the following scenarios:
1. Two socks of one color and one sock each of the other two colors (XXYZ)
WWBG
BBWG
GGWB
2. Three socks of one color and one sock of another color (XXXY)
WWWB
WWWG
BBBW
BBBG
GGGW
GGGB
3. Two socks of each of two colors (XXYY)
WWBB
WWGG
BBGG
4. Four socks of the same color (XXXX)
WWWW
BBBB
GGGG
Calculating the probabilities of each scenario separately would be tedious. Instead, we can find the probability of the complementary event (i.e., selecting four socks with no two socks of the same color) and subtract it from 1. However, it is not possible to draw four socks with no two socks of the same color, since we only have three colors. Therefore, the probability of the complementary event is 0.
Thus, the probability of getting at least two socks of the same color is:
P(at least two of the same color) = 1 - P(complementary event) = 1 - 0 = 1.
Thank you! Also, is the probability of selecting four socks of the same color = ((5/10)*(4/9)*(3/8)*(2/7))+((6/10)*(5/9)*(4/8)*(3/7))+((4/10)*(3/9)*(2/8)*(1/7))?