Problem 952

There are $6$ different
locations at which the path splits, and we label these splits $1$ to $6$, as shown.

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Hide/Reveal Description of Network for Solution 24

There are six different locations where the path splits. The opening leads to split 1.

From split 1, moving to the left leads to split 2 and moving to the right leads to split 4.
From 2, left leads to 3 and right leads to 5.
From 4, left leads to 5 and right leads to 6.
From 3, left leads to bin A and right leads to bin B.
From 5, left leads to bin B and right leads to split 6.
From 6, left leads to bin B and right leads to bin C.

We begin by determining the probability that a ball lands in the bin
labelled $A$.

There is exactly one path that leads to bin $A$.

This path travels downward to the left at each of the three splits
labelled $1$, $2$ and $3$.

At each of these splits, the probability that a ball travels to the left
is $\frac12$, and so the probability
that a ball lands in bin $A$ is $\frac12\times\frac12\times\frac12=\frac18$.

Next, we determine the probability that a ball lands in the bin
labelled $C$.

There are exactly three paths that lead to bin $C$.

One of these paths travels downward to the right at each of the three
splits labelled $1$, $4$ and $6$.

Thus, the probability that a ball lands in bin $C$ by following this path is $\frac12\times\frac12\times\frac12=\frac18$.

A second path to bin $C$ travels
downward to the right at split $1$,
to the left at split $4$, to the
right at split $5$, and to the right
at split $6$.

The probability that a ball follows this path is $\frac12\times\frac12\times\frac12\times\frac12=\frac{1}{16}$.

The third and final path to bin $C$
travels left at split $1$, and to the
right at each of the three splits $2$, $5$
and $6$.

The probability that a ball follows this path is also $\frac{1}{16}$.

The probability that a ball lands in bin $C$ is the sum of the probabilities of
travelling each of these three paths or $$\tfrac{1}{8}+\tfrac{1}{16}+\tfrac{1}{16}=\tfrac{2+1+1}{16}=\tfrac{4}{16}=\tfrac{1}{4}$$

Finally, we determine the probability that a ball lands in bin $B$.

There are six different paths that lead to bin $B$, and we could determine the probability
that a ball follows each of these just as we did for bins $A$ and $C$.

However, it is more efficient to recognize that a ball must land in one
of the three bins, and thus the probability that it lands in bin $B$ is 1 minus the probability that it
lands in bin $A$ minus the
probability that it lands in bin $C$,
or $$1-\tfrac{1}{8}-\tfrac{1}{4}=\tfrac{8-1-2}{8}=\tfrac{5}{8}$$

The probability that the two balls land in different bins is equal to
$1$ minus the probability that the
two balls land in the same bin.

The probability that a ball lands in bin $A$ is $\frac{1}{8}$, and so the probability that
two balls land in bin $A$ is $\frac{1}{8}\times\frac{1}{8}=\frac{1}{64}$.

The probability that a ball lands in bin $C$ is $\frac{1}{4}$, and so the probability that
two balls land in bin $C$ is $\dfrac{1}{4}\times\frac{1}{4}=\frac{1}{16}$.

The probability that a ball lands in bin $B$ is $\frac{5}{8}$, and so the probability that
two balls land in bin $B$ is $\frac{5}{8}\times\frac{5}{8}=\frac{25}{64}$.

Therefore, the probability that the two balls land in different bins is
equal to $$1-\tfrac{1}{64}-\tfrac{1}{16}-\tfrac{25}{64}=\tfrac{64-1-4-25}{64}=\tfrac{34}{64}=\tfrac{17}{32}$$