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I have to calculate the probability of 2 dice and if the result is divisible by 3. The first die will the tens and the second one the units of a two digit number.

E.G. first die i got 5 and second die i got 3. Result: 53 i wrote all number down. i have 36 possible numbers.

My first idea:

The first die doesnt matter. The second die will decide, if the number is divisible by 3 or not.

If my first die is 1. My second die needs to be 2 or 5. If my first die is 2. My second needs to be 1 or 4. If 1st is 3, 2nd needs to be 3 or 6 and so on.

I calculated a probability of 1/3.

Is this correct? I am very bad in formal calculations. Is there a better way to show this "more" mathematical.

FYI: my previous post: Sum of 2 dice rolls is even - probability

hukachaka
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1 Answers1

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An integer is divisible by $3$ if the sum of its digits is divisible by $3$.

If the result of the first throw is a $1$ or $4$, obtaining a $2$ or $5$ on the second throw will yield a two-digit number that is divisible by $3$.

If the result of the first throw is a $2$ or a $5$, obtaining a $1$ or $4$ on the second throw will yield a two-digit number that is divisible by $3$.

If the result of the first throw is a $3$ or $6$, obtaining a $3$ or $6$ on the second throw will yield a two-digit number that is divisible by $3$.

Hence, for any result on the first throw, the probability that you obtain a result on the second throw that results in a number divisible by $3$ is $1/3$ since two of the six possible outcomes are favorable.

N. F. Taussig
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