Three Dice Are Rolled: Find The Probability Of All S

Three Dice Are Rolleda Find The Probability Of Getting All Sixes

Three dice are rolled. (a) Find the probability of getting all sixes. (b) Find the probability of getting all the same outcomes. (c) Find the probability of getting all different outcomes.

Paper For Above instruction

The problem involves calculating probabilities related to rolling three six-sided dice, each with outcomes numbered from 1 to 6. The tasks focus on three specific scenarios: all dice showing sixes, all dice showing the same outcome (but not necessarily six), and all dice showing different outcomes. These probability calculations are grounded in fundamental principles of combinatorics and probability theory, considering the total number of possible outcomes and the favorable outcomes for each scenario.

Part (a): Probability of Getting All Sixes

When rolling three dice, each die can land on any of the six faces, resulting in a total of \(6^3 = 216\) equally likely outcomes. The favorable outcomes for all dice showing sixes occur in only one way: (6, 6, 6). Therefore, the probability \(P\) of getting all sixes is the ratio of favorable outcomes to total outcomes:

\[

P(\text{all sixes}) = \frac{1}{216}

\]

Part (b): Probability of Getting All the Same Outcomes

This scenario considers the probability that all three dice show the same number, which could be 1, 2, 3, 4, 5, or 6. Each such outcome is a triplet like (1, 1, 1), (2, 2, 2), ..., (6, 6, 6). There are 6 favorable outcomes, corresponding to each possible number. Hence, the probability is:

\[

P(\text{all same outcomes}) = \frac{6}{216} = \frac{1}{36}

\]

Part (c): Probability of Getting All Different Outcomes

This scenario involves all three dice showing different numbers. To determine the probability, we first count the number of favorable outcomes where each die shows a unique number, with order mattering since dice are distinguishable.

Number of favorable outcomes:

- For the first die, there are 6 options.

- For the second die, to be different from the first, there are 5 options.

- For the third die, to be different from the first two, there are 4 options.

Thus, total favorable outcomes:

\[

6 \times 5 \times 4 = 120

\]

The total possible outcomes remain 216. Therefore, the probability of all different outcomes is:

\[

P(\text{all different}) = \frac{120}{216} = \frac{20}{36} = \frac{5}{9}

\]

Summary of Results:

- (a) Probability of all sixes: \(\boxed{\frac{1}{216}}\)

- (b) Probability of all the same outcomes: \(\boxed{\frac{1}{36}}\)

- (c) Probability of all different outcomes: \(\boxed{\frac{5}{9}}\)

These calculations illustrate basic probability concepts, including favorable outcomes over total outcomes and combinatorial counting principles relevant to dice probabilities.

References

  • Ross, S. M. (2010). A First Course in Probability (8th ed.). Pearson Education.
  • Grinstead, C. M., & Snell, J. L. (1997). Introduction to Probability. American Mathematical Society.
  • Kreyszig, E. (2011). Probability and Statistics. John Wiley & Sons.
  • Devore, J. L. (2015). Probability and Statistics for Engineering and the Sciences. Cengage Learning.
  • Feller, W. (2008). An Introduction to Probability Theory and Its Applications. Wiley.
  • Snedecor, G. W., & Cochran, W. G. (1989). Statistical Methods. Iowa State University Press.
  • Lehmann, E. L., & Casella, G. (2003). Theory of Point Estimation. Springer.
  • Ross, S. M. (2002). Introduction to Probability Models. Academic Press.
  • Moivre, A. (1738). The Doctrine of Chances. Johnson Reprint Corporation.
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