Question

Kyle challenges you to a game. Each player rolls a standard number cube twice. If the sum of the results is divisible by 3, you win. If it is not, Kyle wins. What is your probability of winning? Is this a fair game?

Answer 1

Answer:

$P(M) = \frac{1}{3}$

The game is not a fair game

Step-by-step explanation:

We start by listing out the sample space

Number Cube 1: {1,2,3,4,5,6}

Number Cube 2: {1,2,3,4,5,6}

Sample Space (S) of the sum of the 2 cubes;

$S = \{2,3,4,5,6,7,3,4,5,6,7,8,4,5,6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12\}$

$n(S) = 36$

Next, is to list out all multiples of 3; Represent this with M

$M = \{3,6,3,6,6,9,6,9,6,9,9,12\}$

$n(M) = 12$

Your probability of winning is the probability of M and is calculated as thus;

$P(M) = \frac{n(M)}{n(S)}$

$P(M) = \frac{12}{36}$

$P(M) = \frac{1}{3}$

The probability of Kyle winning, P(K) = 1 - P(M)

$P(K) = 1 - \frac{1}{3}$

$P(K) = \frac{2}{3}$

The game is not a fair game because;

The probability of you winning and the probability of Kyle winning are not equal;