Hello, simone!

1. The figure shown represents a board with four rows of pegs,

and at the bottom of the board are four cells numbered 1 to 4.

When the ball shown passes between two adjacent pegs in the same row,

it will hit the peg directly beneath the opening.

The ball then has probability ½ of passing to the left or right of that peg.

What is the probability that when the ball will end up in cell 2? Code:

o
↓
* ↓ *
↓
* a *
½/ \½
* b c *
½/ \½ /½
* d e * *
½\ /½
|_______|___o___|_______|_______|
1 2 3 4

There are three paths for the ball to finish in cell 2:

$\displaystyle a-b-d-2:\;P(a,b,d,2) \:=\:\left(\frac{1}{2}\right)^3\:=\:\frac{1}{8} $

$\displaystyle a-b-e-2:\;P(a,b,e,2)\:=\:\left(\frac{1}{2}\right)^3\:=\: \frac{1}{8}$

$\displaystyle a-c-e-2: \;P(a,c,e,2) \:=\:\left(\frac{1}{2}\right)^3 \:=\:\frac{1}{8}$

Therefore: .$\displaystyle P(2) \:=\:3\left(\frac{1}{8}\right) \:=\:\frac{3}{8}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This can also be solved with the Binomial Expansion.

$\displaystyle \left(\frac{1}{2} + \frac{1}{2}\right)^3 \;=\;\underbrace{\left(\frac{1}{2}\right)^3}_{P(1) } + \underbrace{3\left(\frac{1}{2}\right)^2\!\!\left(\ frac{1}{2}\right)}_{P(2)} + \underbrace{3\left(\frac{1}{2}\right)\!\left(\frac {1}{2}\right)^2}_{P(3)} + \underbrace{\left(\frac{1}{2}\right)^3}_{P(4)}$