Question

When two fair 6-sided dice are tossed, the numbers $a$ and $b$ are obtained. What is the probability that the two digit number $ab$ (where $a$ and $b$ are digits) and $a$ and $b$ are all divisible by 3?

Answer 1

Answer: 1/9

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Step-by-step explanation:

If either a or b (or both) are multiples of 3, then so is a*b

Consider 'a' being a multiple of 3. This means a = 3k for some integer k. Multiply both sides by b to get ab = 3k*b, and we see that 3 is a factor of ab, making ab to be a multiple of 3.

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There are six sides to the dice and those sides are {1,2,3,4,5,6} of which only {3,6} are multiples of 3.

If we want both dice to be multiples of 3, then the possible outcomes are

• (3,3)
• (3,6)
• (6,3)
• (6,6)

There are four outcomes. This is out of 6*6 = 36 total outcomes overall.

Dividing the values gives 4/36 = 1/9 as the probability we want

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Side note: for a roll like (a,b) = (6,3) we see that a*b = 6*3 = 18, which is a multiple of 3. The same happens with the other three (a,b) pairings mentioned above.