1. ## probability_please urgent help needed!

thank you.

PROBLEM 1

figure doc.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. Whenever the ball shown passes through the opening between two adjacent pegs in the same row, it will hit the peg directly beneath the opening. The ball then has probability 1/2 of passing through the opening immediately to the left of that peg and probability 1/2 of passing through the opening immediately to the right. What is the probability that when the ball passes throught the first two pegs at the top it will end up in cell 2?

PROBLEM 2

figure doc.2

A person walks from point A to point B along a route confined to the grid of four streets and three avenues shown in the figure. How many routes from A to B can the person take that have minimum possible lenght?

2. 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:

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

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

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

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

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

This can also be solved with the Binomial Expansion.

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

3. thank you so much, really.

could you tell me what you think about the second problem too?

thanks again, congratulations.

4. Originally Posted by simone