Answer :
Certainly! Let's solve the problem of finding the probability of rolling a 1, a 6, and a 4 in any order on three rolls of a fair six-sided die.
1. Determine the probability of rolling each specific number:
Since a fair six-sided die has six faces, the probability of rolling any particular number (such as 1, 6, or 4) on a single roll is:
[tex]\[ P(\text{specific number}) = \frac{1}{6} \][/tex]
2. Identify the probability of rolling the sequence 1, 6, 4:
Since each roll is independent, the probability of rolling the specific sequence of 1, 6, and 4 in that precise order is the product of the individual probabilities:
[tex]\[ P(1 \text{ then } 6 \text{ then } 4) = \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) \][/tex]
3. Calculate the probability for the sequence:
[tex]\[ P(1 \text{ then } 6 \text{ then } 4) = \left(\frac{1}{6}\right)^3 = \frac{1}{216} \][/tex]
So, the probability of rolling the sequence [tex]\(1, 6, 4\)[/tex] on three rolls of a die is approximately [tex]\(0.0046296296296296285\)[/tex].
Here are the intermediate values:
- The probability of each specific number on a single roll: [tex]\(0.16666666666666666\)[/tex]
- The final probability of rolling the sequence [tex]\(1, 6, 4\)[/tex]: [tex]\(0.0046296296296296285\)[/tex]
Thus, [tex]\[P(1, 6, 4) = 0.0046296296296296285\][/tex]
1. Determine the probability of rolling each specific number:
Since a fair six-sided die has six faces, the probability of rolling any particular number (such as 1, 6, or 4) on a single roll is:
[tex]\[ P(\text{specific number}) = \frac{1}{6} \][/tex]
2. Identify the probability of rolling the sequence 1, 6, 4:
Since each roll is independent, the probability of rolling the specific sequence of 1, 6, and 4 in that precise order is the product of the individual probabilities:
[tex]\[ P(1 \text{ then } 6 \text{ then } 4) = \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) \][/tex]
3. Calculate the probability for the sequence:
[tex]\[ P(1 \text{ then } 6 \text{ then } 4) = \left(\frac{1}{6}\right)^3 = \frac{1}{216} \][/tex]
So, the probability of rolling the sequence [tex]\(1, 6, 4\)[/tex] on three rolls of a die is approximately [tex]\(0.0046296296296296285\)[/tex].
Here are the intermediate values:
- The probability of each specific number on a single roll: [tex]\(0.16666666666666666\)[/tex]
- The final probability of rolling the sequence [tex]\(1, 6, 4\)[/tex]: [tex]\(0.0046296296296296285\)[/tex]
Thus, [tex]\[P(1, 6, 4) = 0.0046296296296296285\][/tex]
