What's the probability that we have a box in common?
- Thread starter markallen
- Start date
Hello, markallen!
\(\displaystyle \text{First, we note that }A\text{ can choose }any\text{ four boxes; it doesn't matter.}\)
\(\displaystyle \text{For convenience, let }A\text{ choose boxes }\{X,X,X,X\}\)
. . \(\displaystyle \text{then there are six Others: }\:\{Y,Y,Y,Y,Y,Y\}\)
\(\displaystyle B\text{ chooses 3 of the boxes: }{10\choose3} = 120\text{ possible choices.}\)
\(\displaystyle \text{He wants one of }A\text{'s four boxes: }\:{4\choose1} = 4\text{ ways}\)
. . \(\displaystyle \text{and 2 of the 6 Others: }\:{6\choose2} = 15\text{ ways.}\)
\(\displaystyle \text{Hence, there are: }\:4\cdot15 \,=\,60\text{ ways B can have one box in common.}\)
\(\displaystyle \text{Therefore: }\
(\text{1 box in common}) \:=\:\frac{60}{120} \:=\:\frac{1}{2}\)
\(\displaystyle B\text{ wants two of }A\text{'s four boxes: }\:{4\choose2} = 6\text{ ways}\)
, . \(\displaystyle \text{and one of the 6 Others: }\:{6\choose1} = 6\text{ ways.}\)
\(\displaystyle \text{Hence, there are: }\:6\cdot6 = 36\text{ ways }B\text{ can have two boxes in common.}\)
\(\displaystyle \text{Therefore: }\(\text{two boxes in common}) \:=\:\frac{36}{120} \:=\:\frac{2}{5}\)
\(\displaystyle B\text{ wants three of }A\text{'s four boxes: }\:{4\choose 3} = 4\text{ ways.}\)
\(\displaystyle \text{Hence, there are: }\:4\text{ ways }B\text{ can have three boxes in common.}\)
\(\displaystyle \text{Therefore: }\(\text{three boxes in common}) \:=\:\frac{4}{120} \:=\:\frac{1}{30}\)
\(\displaystyle \text{First, we note that }A\text{ can choose }any\text{ four boxes; it doesn't matter.}\)
\(\displaystyle \text{For convenience, let }A\text{ choose boxes }\{X,X,X,X\}\)
. . \(\displaystyle \text{then there are six Others: }\:\{Y,Y,Y,Y,Y,Y\}\)
\(\displaystyle B\text{ chooses 3 of the boxes: }{10\choose3} = 120\text{ possible choices.}\)
\(\displaystyle \text{He wants one of }A\text{'s four boxes: }\:{4\choose1} = 4\text{ ways}\)
. . \(\displaystyle \text{and 2 of the 6 Others: }\:{6\choose2} = 15\text{ ways.}\)
\(\displaystyle \text{Hence, there are: }\:4\cdot15 \,=\,60\text{ ways B can have one box in common.}\)
\(\displaystyle \text{Therefore: }\
(\text{1 box in common}) \:=\:\frac{60}{120} \:=\:\frac{1}{2}\)\(\displaystyle B\text{ wants two of }A\text{'s four boxes: }\:{4\choose2} = 6\text{ ways}\)
, . \(\displaystyle \text{and one of the 6 Others: }\:{6\choose1} = 6\text{ ways.}\)
\(\displaystyle \text{Hence, there are: }\:6\cdot6 = 36\text{ ways }B\text{ can have two boxes in common.}\)
\(\displaystyle \text{Therefore: }\
(\text{two boxes in common}) \:=\:\frac{36}{120} \:=\:\frac{2}{5}\)\(\displaystyle B\text{ wants three of }A\text{'s four boxes: }\:{4\choose 3} = 4\text{ ways.}\)
\(\displaystyle \text{Hence, there are: }\:4\text{ ways }B\text{ can have three boxes in common.}\)
\(\displaystyle \text{Therefore: }\
(\text{three boxes in common}) \:=\:\frac{4}{120} \:=\:\frac{1}{30}\)