Question

Consider the following scenario. Bob is playing a board game where he blindly picks a tile from a container and moves some spaces. The container has 5 red tiles, 11 green tiles, and 4 blue tiles. If he draws a red tile, he moves forward 12 spaces. If he draws a green tile, he stays put. If he draws a blue tile, he moves backward 15 spaces. Use mathematical calculations to show whether or not the game is fair. That is, show whether Bob will hover around the same space (fair) or, over time, move away from the starting point (not fair). Show all of your calculations for full credit.

Answer 1

Answer:

Yes, the game is fair.

Step-by-step explanation:

In order to check that the game is fair we will see that the total expectation is zero.

i.e. Bob will hover around the same place.

There are total 20 tiles in a container.

( Since 5 red tile+11 green tile + 4 blue tile=20 tiles)

The probability of choosing a red tile is:

    Number of red tile/total number of tiles

     =  5/20

Also, the probability of choosing a green tile is:

   Number of green tile/Total number of tiles

 = 11/20

Also, the probability of choosing a blue tile is:

   Number of blue tiles/ Total number of tiles

     =  4/20

Let E denote the expectation of an event.

Hence,

$E(Red\ tile)=\dfrac{5}{20}\times (+12)$

Since when a red tile is drawn Bob moves 12 spaces forward.

$E( Red\ tile)=\dfrac{60}{20}$

Similarly

$E(Green\ tile)=\dfrac{11}{20}\times (0)$

Since when a green tile is drawn Bob remains at the same place.

$E( Green\ tile)=0$

and

$E(Blue\ tile)=\dfrac{4}{20}\times (-15)$

Since when a blue tile is drawn Bob moves 15 spaces backward

$E(Blue\ tile)=\dfrac{-60}{20}$

Hence, total expectation is:

$=E(Red\ tile)+E(Green\ tile)+E(Blue\ tile)\\\\\\=\dfrac{60}{20}+0-\dfrac{60}{20}\\\\=0$

       This means that he will around the same place.

             Hence, the game is fair.

Answer 2

The game is fair. 

Pls. see my attachment. 

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