Question

When a certain biased die is rolled, a particular face occurs with probability $\frac{1}{6} - x$ and its opposite face occurs with probability $\frac{1}{6} + x$. All other faces occur with probability $\frac{1}{6}$. Note that opposite faces sum to 7 in any die. If $0<x<\frac{1}{6}$, and the probability of obtaining total sum $= 7$, when such a die is rolled twice, is $\frac{13}{96}$, then the value of $x$ is :

• A. $\frac{1}{9}$
• B. $\frac{1}{16}$
• C. $\frac{1}{12}$
• D. $\frac{1}{8}$

Hint:

Notice that the opposite faces always appear in pairs $(a, b)$ and $(b, a)$ in the sum for 7. For unbiased dice, this sum is $6 \times (1/36) = 1/6$. Here, only one pair is biased, allowing for a quick setup.

Answer 1

The Correct Option is D

The problem involves calculating the probability that the sum of the numbers on two biased dice equals 7. Given:

• One biased face with probability \(\frac{1}{6} - x\)
• The opposite face with probability \(\frac{1}{6} + x\)
• All other faces with probability \(\frac{1}{6}\)

We know that opposite faces sum to 7, which means if one face shows 1, the opposite shows 6, and so on. These are the face pairs:

• (1,6), (2,5), (3,4)

The sum of the numbers showing on the top when two dice are rolled to get a total of 7 could be formed by:

• (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)

The probability of getting each pair is due to the combinations of the biased and unbiased probabilities. Let's calculate the probability of each pair contributing to a sum of 7:

1. The events (1,6) and (6,1):
   • Probability (1,6) is \((\frac{1}{6} - x)(\frac{1}{6} + x)\)
   • Probability (6,1) is the same: \((\frac{1}{6} + x)(\frac{1}{6} - x)\)
2. The events (2,5) and (5,2):
   • Probability (2,5) is \((\frac{1}{6})(\frac{1}{6})\)
   • Probability (5,2) is \((\frac{1}{6})(\frac{1}{6})\)
3. The events (3,4) and (4,3):
   • Probability (3,4) is \((\frac{1}{6})(\frac{1}{6})\)
   • Probability (4,3) is \((\frac{1}{6})(\frac{1}{6})\)

Now, summing these probabilities, we note:

• Probability for (1,6) and (6,1): 2\left( \frac{1}{6} - x \right)\left( \frac{1}{6} + x \right) = 2\left( \frac{1}{36} - x^2 \right)
• Probability for (2,5), (5,2), (3,4), (4,3): 4 \times \left(\frac{1}{6} \times \frac{1}{6}\right) = \frac{4}{36}

Therefore, the total probability is 2\left( \frac{1}{36} - x^2 \right) + \frac{4}{36} = \frac{13}{96}.

Solving for \(x\):

• Simplify: \frac{2}{36} - 2x^2 + \frac{4}{36} = \frac{13}{96}
• Convert all fractions to a common denominator and solve for \(x\):
• On simplification: 2\left(\frac{1}{36}\right) + \frac{4}{36} - 2x^2 = \frac{13}{96}
• \frac{6}{36} - 2x^2 = \frac{13}{96}
• Solve for \(x^2\): equivalent to solving \frac{1}{6} - 2x^2 = \frac{13}{96}
• After calculations: x^2 = \frac{1}{64}, \(\therefore x = \frac{1}{8}\).

Hence, the value of \(x\) is \(\frac{1}{8}\).