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:
 \:=\:\left(\frac{1}{2}\right)^3\:=\:\frac{1}{8} )
\:=\:\left(\frac{1}{2}\right)^3\:=\: \frac{1}{8})
 \:=\:\left(\frac{1}{2}\right)^3 \:=\:\frac{1}{8})
Therefore: .  \:=\:3\left(\frac{1}{8}\right) \:=\:\frac{3}{8})
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
This can also be solved with the Binomial Expansion.
^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)})
LinkBack URL
About LinkBacks
Originally Posted by simone 
