Math Problem Statement

A six-faced die is so biased that it is twice as likely to show an even number as an odd number. It is tossed twice. Find the probability that the sum of the numbers obtained shall be even

Solution

To solve the problem, we first analyze the bias of the die and the outcomes of each roll.

Step 1: Determine Probabilities for Each Roll

The die is biased such that even numbers are twice as likely to appear as odd numbers. Let's assign probabilities based on this information.

  • Let the probability of rolling an odd number (1, 3, or 5) be pp.
  • Then, the probability of rolling an even number (2, 4, or 6) is 2p2p.

Since there are 3 odd numbers and 3 even numbers, we must have:

3p+3(2p)=13p + 3(2p) = 1

3p+6p=13p + 6p = 1

9p=19p = 1

p=19p = \frac{1}{9}

Thus, the probability of rolling an odd number is 19\frac{1}{9}, and the probability of rolling an even number is 29\frac{2}{9}.

Step 2: Calculate the Probability that the Sum is Even

The sum of two numbers is even if both numbers are either even or odd. We need to calculate the probability of these two cases:

  1. Both rolls are even:

    • Probability of the first roll being even: 23\frac{2}{3}
    • Probability of the second roll being even: 23\frac{2}{3}
    • Combined probability: 23×23=49\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}
  2. Both rolls are odd:

    • Probability of the first roll being odd: 13\frac{1}{3}
    • Probability of the second roll being odd: 13\frac{1}{3}
    • Combined probability: 13×13=19\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}

Step 3: Add the Probabilities

To find the total probability that the sum is even, we add the probabilities of both scenarios:

Total probability=49+19=59\text{Total probability} = \frac{4}{9} + \frac{1}{9} = \frac{5}{9}

Final Answer:

The probability that the sum of the numbers obtained is even is 59\frac{5}{9}.


Would you like more details or have any questions?

Here are 5 related questions to consider:

  1. What is the probability of obtaining a sum that is odd in this scenario?
  2. How would the solution change if the die were fair?
  3. What if the die were biased three times in favor of even numbers?
  4. How do we extend this to three rolls and find the probability that the sum is even?
  5. How would you calculate the expected value of the sum of two rolls?

Tip: Always break down biased probability problems by first determining individual probabilities before combining them for the final outcome.

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Math Problem Analysis

Mathematical Concepts

Probability
Conditional Probability

Formulas

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Theorems

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Suitable Grade Level

Grades 10-12