Math Problem Statement

What is the probability of rolling a die and landing on 1, 4, or 6 on the 2nd, 3rd, or 4th roll?

Solution

The task involves determining the probability of rolling a die and landing on 1, 4, or 6 on the 2nd, 3rd, or 4th roll. Let me break this down step by step.


Analysis of the Problem:

  1. Probability of Success (rolling a 1, 4, or 6):

    • A die has 6 faces, and there are 3 favorable outcomes (1, 4, 6).
      Psuccess=36=0.5P_{\text{success}} = \frac{3}{6} = 0.5
  2. Probability of Failure (not rolling 1, 4, or 6):

    • The complementary probability is: Pfailure=1Psuccess=10.5=0.5P_{\text{failure}} = 1 - P_{\text{success}} = 1 - 0.5 = 0.5
  3. Probability of Success on Specific Rolls:

    • The event happens on the 2nd, 3rd, or 4th roll.
      For the 2nd roll, we must fail on the 1st roll and succeed on the 2nd.
      Similarly, for the 3rd or 4th roll: Psuccess at 2nd roll=PfailurePsuccess=0.50.5=0.25P_{\text{success at 2nd roll}} = P_{\text{failure}} \cdot P_{\text{success}} = 0.5 \cdot 0.5 = 0.25 Psuccess at 3rd roll=Pfailure2Psuccess=0.520.5=0.125P_{\text{success at 3rd roll}} = P_{\text{failure}}^2 \cdot P_{\text{success}} = 0.5^2 \cdot 0.5 = 0.125 Psuccess at 4th roll=Pfailure3Psuccess=0.530.5=0.0625P_{\text{success at 4th roll}} = P_{\text{failure}}^3 \cdot P_{\text{success}} = 0.5^3 \cdot 0.5 = 0.0625
  4. Total Probability of Success at 2nd, 3rd, or 4th Roll:

    • The events are mutually exclusive, so we sum the probabilities: Ptotal=Psuccess at 2nd+Psuccess at 3rd+Psuccess at 4thP_{\text{total}} = P_{\text{success at 2nd}} + P_{\text{success at 3rd}} + P_{\text{success at 4th}} Ptotal=0.25+0.125+0.0625=0.4375P_{\text{total}} = 0.25 + 0.125 + 0.0625 = 0.4375
  5. Final Answer:

    • The probability of success at the 2nd, 3rd, or 4th roll is: 0.4375\boxed{0.4375}

Would you like me to explain any specific steps or fill in the missing fields in the solution image?


Questions to Explore Further:

  1. What is the probability of success if the favorable outcomes are changed (e.g., 1, 2, 3)?
  2. How does the calculation change if the die is biased (not fair)?
  3. Can this be modeled using a geometric distribution? If so, how?
  4. What if we include the 1st roll or exclude the 4th roll in the success conditions?
  5. What is the general formula for probabilities across multiple independent events?

Tip:

When solving problems involving probabilities across multiple trials, visualize the failure and success paths as a tree to ensure clarity and correctness.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Independent Events

Formulas

Probability of an event: P(A) = Number of favorable outcomes / Total number of possible outcomes
Probability of independent events: P(A and B) = P(A) * P(B)

Theorems

Multiplication Rule for Independent Events

Suitable Grade Level

Grades 9-12