Probability that given weight is on the heavier side of a balance

Assume we have 6 balls weighting 101, 102, 103, 104, 105 and 106 in kg. If we randomly pick 3 balls and place them on one side of the scale and the other 3 left are placed on the other side of the scale. What is the probability that the ball weighting 106 kg is on the heavier side?

Solution 1:

Let $X,Y$ be the other two balls on the same side as the $106kg$ ball,

The event "106kg is on the heavier side"($E$) can be expressed as the following:

$$E \iff X+Y+106 \geq \left \lceil \frac{621}{2} \right\rceil = 311 \implies X+ Y \geq 311 - 106 = 205 \implies $$

$$P(E) = P(X + Y \geq 205)$$

If we set aside 106 then $X,Y$ are chosen from the remaining 5 balls, so there are ${5 \choose 2} = 10$ total possibilities. Of these, the ones that result in $205$ or higher are:

$(101,104),(101,105), (102,103), (102,104), (102,105), (103,104),(103,105),(104,105)$

So 8 possibilities out of 10, so $80\%$ probability that the side that htas 106 kg ball is the heavier side..

Solution 2:

The problem is same as considering the weights $1,2,3,4,5,6$ instead.

There are $\dfrac{6!}{3!3!2!}=10$ ways to divide $6$ distinct objects into two groups of threes. One way, quick enough, is to list out the ten pairs of triples and see for how many cases $6$ belongs to the group with larger sum.

A yet shorter way is to consider the sum $1+2+3+4+5+6=21$. $6$ belongs to heavier side, when sum of three weights including $6$ is more than $21/2=10.5$. Unfavorable cases are only two : $(6,1,2)$ and $(6,1,3)$.

Number of favorable events is $10-2=8$ and required probability is $8/10$.