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

Step 1: Understanding the Concept:
The probability of a combined event is the sum of probabilities of mutually exclusive outcomes. We identify all pairs of outcomes $(d_1, d_2)$ that sum to 7.
Step 2: Detailed Explanation:
Let the biased faces be 1 and 6 (as $1+6=7$).
$P(1) = \frac{1}{6} - x$, $P(6) = \frac{1}{6} + x$.
Other faces: $P(2) = P(3) = P(4) = P(5) = \frac{1}{6}$.
The pairs $(d_1, d_2)$ summing to 7 are: $(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$.
Probability of sum 7 is:
\[ P(\text{sum}=7) = P(1)P(6) + P(6)P(1) + P(2)P(5) + P(5)P(2) + P(3)P(4) + P(4)P(3) \]
\[ \frac{13}{96} = 2 \left( \frac{1}{6}-x \right) \left( \frac{1}{6}+x \right) + 2 \left( \frac{1}{6} \right) \left( \frac{1}{6} \right) + 2 \left( \frac{1}{6} \right) \left( \frac{1}{6} \right) \]
\[ \frac{13}{96} = 2 \left( \frac{1}{36} - x^2 \right) + \frac{2}{36} + \frac{2}{36} \]
\[ \frac{13}{96} = \frac{2}{36} - 2x^2 + \frac{4}{36} = \frac{6}{36} - 2x^2 = \frac{1}{6} - 2x^2 \]
\[ 2x^2 = \frac{1}{6} - \frac{13}{96} = \frac{16 - 13}{96} = \frac{3}{96} = \frac{1}{32} \]
\[ x^2 = \frac{1}{64} \implies x = \frac{1}{8} \quad (\text{since } x>0) \]
Step 3: Final Answer:
The value of $x$ is $1/8$.