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Step-by-Step Solution
Step 1: Identify the scenario
A biased die is rolled so that one particular face appears with probability $ \frac{1}{6} - x $, its opposite face with probability $ \frac{1}{6} + x $, and each of the remaining four faces with probability $ \frac{1}{6} $. We roll this biased die twice and want the probability of obtaining a total sum of 7 to be $ \frac{13}{96} $.
Step 2: List the pairs that sum to 7
The pairs of faces on two dice that add up to 7 are:
(1, 6) and (6, 1)
(2, 5) and (5, 2)
(3, 4) and (4, 3)
Step 3: Determine the probabilities of each pair
Assume the “particular face” is 1 (with probability $ \frac{1}{6} - x $) and its opposite face is 6 (with probability $ \frac{1}{6} + x $). Then:
Probability(1 on first die and 6 on second die)
= $ \left(\frac{1}{6} - x\right) \left(\frac{1}{6} + x\right) $.
Probability(6 on first die and 1 on second die)
= $ \left(\frac{1}{6} + x\right) \left(\frac{1}{6} - x\right) $.
Probability(2 on one die and 5 on the other)
= $ \frac{1}{6} \times \frac{1}{6} $ (since 2 and 5 each appear with probability $ \frac{1}{6} $).
Probability(3 on one die and 4 on the other)
= $ \frac{1}{6} \times \frac{1}{6} $ similarly.
Step 4: Sum up all favorable outcomes
All pairs (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) constitute the event “sum = 7.” Combine these probabilities:
$
\displaystyle
\text{Probability(sum = 7)}
= \left(\frac{1}{6} - x\right)\left(\frac{1}{6} + x\right)
+ \left(\frac{1}{6} + x\right)\left(\frac{1}{6} - x\right)
+ \frac{1}{6}\cdot \frac{1}{6} + \frac{1}{6}\cdot \frac{1}{6} + \frac{1}{6}\cdot \frac{1}{6} + \frac{1}{6}\cdot \frac{1}{6}.
$
However, to make it simpler, notice there are 3 distinct pairs, each with two permutations. So we can factor out a 2 for the symmetrical cases (1,6) & (6,1), (2,5) & (5,2), (3,4) & (4,3). Thus a shorter expression used in the given solution is:
$
\displaystyle
2 \Bigl[ \left(\tfrac{1}{6} - x\right)\left(\tfrac{1}{6} + x\right) \;+\; \tfrac{1}{6}\cdot\tfrac{1}{6} \;+\; \tfrac{1}{6}\cdot\tfrac{1}{6} \Bigr]
= \frac{13}{96}.
$
Step 5: Solve for x
First expand $ \left(\frac{1}{6} - x\right)\left(\frac{1}{6} + x\right) $:
$
\left(\frac{1}{6} - x\right)\left(\frac{1}{6} + x\right)
= \left(\frac{1}{6}\right)^2 - x^2
= \frac{1}{36} - x^2.
$
Hence inside the brackets we have:
$
\frac{1}{36} - x^2 \;+\; \frac{1}{36} \;+\; \frac{1}{36}
= \frac{3}{36} - x^2
= \frac{1}{12} - x^2.
$
Thus the total probability is:
$
2\Bigl(\tfrac{1}{12} - x^2\Bigr)
= \frac{13}{96}.
$
Simplify:
$
\tfrac{1}{12} - x^2 = \tfrac{13}{192}
\quad\Longrightarrow\quad
x^2 = \tfrac{1}{12} - \tfrac{13}{192}.
$
$
\tfrac{1}{12} = \tfrac{16}{192}, \quad
\tfrac{16}{192} - \tfrac{13}{192} = \tfrac{3}{192} = \tfrac{1}{64}.
$
$
x^2 = \tfrac{1}{64}
\quad\Longrightarrow\quad
x = \tfrac{1}{8} \quad (\text{taking the positive root since } 0
