What's the probability that we have a box in common?

[markallen]

Suppose there are 10 identical boxes.

If person A chooses 4 boxes at random, and person B chooses 3 boxes at random, what is the probability that:

a) They have 1 box in common
b) They have 2 boxes in common
c) They have 3 boxes in common

 

[pka]

Suppose there are 10 identical boxes.
If person A chooses 4 boxes at random, and person B chooses 3 boxes at random, what is the probability that:
a) They have 1 box in common
b) They have 2 boxes in common
c) They have 3 boxes in common

Does that make any sense?
If the boxes are identical how would we know any difference?

 

[Denis]

If person A chooses 4 boxes at random, and person B chooses 3 boxes at random, what is the probability that:
a) They have 1 box in common

Where's your work on this? :shock:

Label the boxes 1 to 10 : see why?
HINT: after A chooses his 4 boxes, B has 4/10 chance of choosing one of A's : understand?

 

[markallen]

Yea I get that part, which a very handy tip.

But when it comes to actually answering the question, i.e. when Person B picks 3 boxes... I can't quite get my head round the calculation.

 

[soroban]

Hello, markallen!

Suppose there are 10 identical boxes.

If person A chooses 4 boxes at random, and person B chooses 3 boxes at random,
what is the probability that:

a) They have 1 box in common

\(\displaystyle \text{First, we note that }A\text{ can choose }any\text{ four boxes; it doesn't matter.}\)

\(\displaystyle \text{For convenience, let }A\text{ choose boxes }\{X,X,X,X\}\)
. . \(\displaystyle \text{then there are six Others: }\:\{Y,Y,Y,Y,Y,Y\}\)

\(\displaystyle B\text{ chooses 3 of the boxes: }{10\choose3} = 120\text{ possible choices.}\)

\(\displaystyle \text{He wants one of }A\text{'s four boxes: }\:{4\choose1} = 4\text{ ways}\)
. . \(\displaystyle \text{and 2 of the 6 Others: }\:{6\choose2} = 15\text{ ways.}\)

\(\displaystyle \text{Hence, there are: }\:4\cdot15 \,=\,60\text{ ways B can have one box in common.}\)

\(\displaystyle \text{Therefore: }\(\text{1 box in common}) \:=\:\frac{60}{120} \:=\:\frac{1}{2}\)

b) They have 2 boxes in common

\(\displaystyle B\text{ wants two of }A\text{'s four boxes: }\:{4\choose2} = 6\text{ ways}\)
, . \(\displaystyle \text{and one of the 6 Others: }\:{6\choose1} = 6\text{ ways.}\)

\(\displaystyle \text{Hence, there are: }\:6\cdot6 = 36\text{ ways }B\text{ can have two boxes in common.}\)

\(\displaystyle \text{Therefore: }\(\text{two boxes in common}) \:=\:\frac{36}{120} \:=\:\frac{2}{5}\)

c) They have 3 boxes in common

\(\displaystyle B\text{ wants three of }A\text{'s four boxes: }\:{4\choose 3} = 4\text{ ways.}\)

\(\displaystyle \text{Hence, there are: }\:4\text{ ways }B\text{ can have three boxes in common.}\)

\(\displaystyle \text{Therefore: }\(\text{three boxes in common}) \:=\:\frac{4}{120} \:=\:\frac{1}{30}\)

 

[Denis]

OR:
if the X box picked first: (4/10) * (6/9) * (5/8) = 1/6
Since 2 other ways to pick X box (2nd and 3rd) then: 3 * (1/6) = 1/2