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

IMLOG10ct 2021-11-21 Answered
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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Expert Answer

inenge3y
Answered 2021-11-22 Author has 591 answers
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 \(\displaystyle={x}={\frac{{{1}}}{{{16}}}}\)
Answer:
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) \(\displaystyle{\frac{{{1}}}{{{16}}}}\)
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