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baherus [9]
3 years ago
12

Consider rolling two fair dice one 3-sided the other 5-sided

Mathematics
1 answer:
Ne4ueva [31]3 years ago
8 0

Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose

1\leq x\leq 3,\quad 1\leq y \leq 5

We have P(X=x)=\frac{1}{3} and P(Y=y)=\frac{1}{5}, because the dice are fair.

Now we use the assumption of independence to claim that

P(X=x, Y=y) = P(X=x)\cdot P(Y=y) =\dfrac{1}{3}\cdot\dfrac{1}{5} = \dfrac{1}{15}

Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:

  • 2 in a unique way (1+1)
  • 3 in two possible ways (1+2, 2+1)
  • 4 in three possible ways
  • 5 in three possible ways
  • 6 in three possible ways
  • 7 in two possible ways
  • 8 in a unique way

This implies that the probabilities of the outcomes of W=X+Y are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5

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