# Suppose a die is weighted such that the probability of rolling a three

Suppose a die is weighted such that the probability of rolling a three is the same as rolling a six, the probability of rolling a one, two, or four is 3 times that of six, and the probability of rolling a five is 5 times that of rolling a three. Find the probability of 1. rolling a one 2. rolling a two 3. rolling a three 4. rolling a four 5. rolling a five 6. rolling a six

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inenge3y
Step 1
Let the probability of rolling a six is $$\displaystyle={x}$$
So the probability of rolling a three is $$\displaystyle={x}$$
the probability of rolling a one is $$\displaystyle={3}{x}$$
the probability of rolling a two is $$\displaystyle={3}{x}$$
the probability of rolling a four is $$\displaystyle={3}{x}$$
the probability of rolling a five is $$\displaystyle={5}{x}$$
Step 2
We know that the total probability is 1.
So, $$\displaystyle{x}+{x}+{3}{x}+{3}{x}+{3}{x}+{5}{x}={1}$$
Or, $$\displaystyle{16}{x}={1}$$
Or, $$\displaystyle{x}={\frac{{{1}}}{{{16}}}}$$
The probability of rolling a one is $$\displaystyle={3}{x}={3}\times{\frac{{{1}}}{{{16}}}}={\frac{{{3}}}{{{16}}}}$$
The probability of rolling a two is $$\displaystyle={3}{x}={3}\times{\frac{{{1}}}{{{16}}}}={\frac{{{3}}}{{{16}}}}$$
The probability of rolling a three is $$\displaystyle={x}={\frac{{{1}}}{{{16}}}}$$
The probability of rolling a four is $$\displaystyle={3}{x}={3}\times{\frac{{{1}}}{{{16}}}}={\frac{{{3}}}{{{16}}}}$$
The probability of rolling a five is $$\displaystyle={5}{x}={5}\times{\frac{{{1}}}{{{16}}}}={\frac{{{5}}}{{{16}}}}$$
The probability of rolling a six is $$\displaystyle={x}={\frac{{{1}}}{{{16}}}}$$
1) $$\displaystyle{\frac{{{3}}}{{{16}}}}$$
2) $$\displaystyle{\frac{{{3}}}{{{16}}}}$$
3) $$\displaystyle{\frac{{{1}}}{{{16}}}}$$
4) $$\displaystyle{\frac{{{3}}}{{{16}}}}$$
5) $$\displaystyle{\frac{{{5}}}{{{16}}}}$$
6) $$\displaystyle{\frac{{{1}}}{{{16}}}}$$