Why Does the Method of Distribution Affect Probability Calculations?

[mooncrater]

Homework Statement

The question says that :
Twelve distinct balls are distributes among three distinct boxes. Find the probability that the first box contains 3 balls.

Homework Equations

None

The Attempt at a Solution

It can be done through:
No. Of choices for each ball=3
Total no. of choices=3¹²
We have to select 3 out of twelve and arrange the remaining nine balls in 2⁹ ways so that the answer would be :
110×2¹⁰/3¹².
But I was thinking that why a choice is given to the balls. Why not to boxes? Each box then will have 12 choices and total choices would be 12³. Why this thinking is incorrect here?

 

[paisiello2]

The first way is you are going to assign each ball to one of the 3 boxes until all the balls are used up. No balls are left over.

The second way is you are attempting to assign each box to one of 12 balls until all the boxes are used up. Since the number of boxes is less than the number of balls then there will be 9 balls left over without any boxes assigned. You also calculated it wrong since you are assuming the balls would be re-used. The correct calculation should be 12⋅11⋅10 = 13,200.

 

[haruspex]

Each box then will have 12 choices

But those choices are not independent. Two boxes cannot choose the same ball, but two balls can choose the same box.

You also calculated it wrong since you are assuming the balls would be re-used. The correct calculation should be 12⋅11⋅10 = 13,200.

mooncrater's answer looks ok to me. ¹²C₃(1/3)³(2/3)⁹.
One thing bothers me... why are we told the balls are distinct?

balls are distributes among three distinct boxes

That's a bit vague. I'd rather it said "each ball is assigned independently and with equal probability to any of three boxes".

 

[mooncrater]

But those choices are not independent. Two boxes cannot choose the same ball, but two balls can choose the same box.

Ok got it now. Thanks.

 

[Ray Vickson]

Homework Statement

The question says that :
Twelve distinct balls are distributes among three distinct boxes. Find the probability that the first box contains 3 balls.

Homework Equations

None

The Attempt at a Solution

It can be done through:
No. Of choices for each ball=3
Total no. of choices=3¹²
We have to select 3 out of twelve and arrange the remaining nine balls in 2⁹ ways so that the answer would be :
110×2¹⁰/3¹².
But I was thinking that why a choice is given to the balls. Why not to boxes? Each box then will have 12 choices and total choices would be 12³. Why this thinking is incorrect here?

When assigning balls to boxes, the box choices are independent from ball-to-ball, so you have a binomial distribution with n = 12 and the two categories ('box 1' or 'not box 1'). The probability that box 1 contains $k$ balls is
$$P(k) = {12 \choose k} \, (1/3)^k \, (2/3)^{12-k}, \; k = 0,1,2, \ldots, 12.$$
Substitute $k = 3$

 

[paisiello2]

mooncrater's answer looks ok to me. ¹²C₃(1/3)³(2/3)⁹.

You're confused about which number I was referring to. He calculated the total number of choices for the box as 12³ which can only be true if the balls are replaced at every choice. Which is another different assumption than when choosing for the balls.