Question

game a and game b are played by tossing two 6 - sided number cubes. the numbers on the cubes determine which player gets a point, as shown in the rules below. game a: • player 1 gets a point if the product of the numbers is even. • player 2 gets a point if the product is odd. game b: • player 1 gets a point if the sum of the numbers is even. • player 2 gets a point if the sum is odd. are game a and game b fair for the two players? game a is fair, and game b is not fair. both game a and game b are fair. game a is not fair, and game b is fair. neither game a nor game b is fair.

Explanation:

Step1: Analyze Game A (Product)

A product is odd only when both numbers are odd. For a 6 - sided cube, odd numbers are 1, 3, 5 (3 odd numbers) and even numbers are 2, 4, 6 (3 even numbers). The probability that the first cube is odd is $\frac{3}{6}=\frac{1}{2}$, and the same for the second cube. So the probability that the product is odd (Player 2 gets a point) is $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$? Wait, no. Wait, the number of possible outcomes when tossing two 6 - sided cubes is $n(S)=6\times6 = 36$. The number of outcomes where both are odd: for the first cube, 3 choices (1,3,5), for the second cube, 3 choices. So number of favorable outcomes for odd product is $3\times3 = 9$. So probability of odd product (Player 2) is $\frac{9}{36}=\frac{1}{4}$? No, wait, $3\times3 = 9$, $36$ total. Then probability of even product (Player 1) is $1-\frac{9}{36}=\frac{27}{36}=\frac{3}{4}$. Wait, that can't be. Wait, no: a product is even if at least one number is even. The number of outcomes where at least one is even is total outcomes minus both odd. So $36 - 9=27$. So Player 1 has probability $\frac{27}{36}=\frac{3}{4}$, Player 2 has $\frac{9}{36}=\frac{1}{4}$. So Game A is not fair? Wait, no, wait I made a mistake. Wait, odd numbers on a die: 1,3,5 (3 numbers), even: 2,4,6 (3 numbers). The product of two numbers is odd if and only if both numbers are odd. So the number of ways to get two odd numbers: for the first die, 3 choices, second die, 3 choices, so $3\times3 = 9$ outcomes. The number of ways to get an even product: total outcomes (36) minus 9 = 27. So probability Player 1 (even product) gets a point is $\frac{27}{36}=\frac{3}{4}$, Player 2 (odd product) is $\frac{9}{36}=\frac{1}{4}$. So Game A is not fair.

Step2: Analyze Game B (Sum)

A sum is even if both numbers are even or both are odd. Let's calculate the number of outcomes. Case 1: both odd. Number of outcomes: $3\times3 = 9$. Case 2: both even. Number of outcomes: $3\times3 = 9$. So total number of outcomes where sum is even (Player 1) is $9 + 9=18$. The number of outcomes where sum is odd (Player 2) is total outcomes (36) minus 18 = 18. So probability Player 1 gets a point is $\frac{18}{36}=\frac{1}{2}$, and Player 2 also $\frac{18}{36}=\frac{1}{2}$. So Game B is fair.

So Game A is not fair, Game B is fair.

Answer:

Game A is not fair, and Game B is fair.