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You are playing a dice game. The object of the game is to roll a certain combination of numbers in a single toss of two dice. Your dice are 'fair', but your opponent likes to win, and his dice are 'loaded' such that a four has a 7/36 probability of landing upwards. That means that the opposite side of four is counter-loaded (the opposite sides of a die adds to seven). What is the probability of your opponent rolling two even numbers? Give your answer as a decimal value, not a percent.

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Answer:

There is a 0.2785 probability of your opponent rolling two even numbers.

Step-by-step explanation:

Since we are looking for the probability of rolling even numbers, we must consider the odds of rolling a 2, a 4 or a 6.

A fair dice would yield in a 6/36 chance of rolling each number, but since these dice are loaded, there is a 7/36 chance of rolling a 4 and a 5/36 chance of rolling a 3 (opposite side of 4); the probabilities for rest remain the same.

Therefore, the probability (P1) of rolling and even number in one die is:

[tex]P_{1} = P(2) + P (4) + P (6)\\P_{1} = \frac{6}{36} + \frac{7}{36} +\frac{6}{36} \\P_{1} = \frac{19}{36}[/tex]

Since the game involves two dice, the probability of rolling even numbers on both is:

[tex]P_{2} = P_{1} ^{2}\\P_{2} = (\frac{19}{36}) ^{2}\\P_{2} =0.2785[/tex]

There is a 0.2785 probability of your opponent rolling two even numbers.

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