In a certain biased die, the probability of getting a particular face is (1/6 + x) and the opposite is (1/6 - x) and 0 < x < 1/6. The probability for other faces is 1/6 each. The sum of opposite faces in a die is 7. If the probability of getting a sum 7 in a throw of 2 such die is 13/96, then x is
(a) \(\frac{1}{8}\)
(b) \(\frac{1}{12}\)
(c) \(\frac{1}{18}\)
(d) \(\frac{1}{20}\)
Correct answer: (c) \(\frac{1}{18}\)
Explanation:
For particular and opposite faces, if sum is 7.
Probability = 2(\(\frac{1}{6}\) + x)(\(\frac{1}{6}\) - x)
For remaining faces in two die, if sum is 7,
Probability = 2(\(\frac{1}{6}\) x \(\frac{1}{6}\) + \(\frac{1}{6}\) x \(\frac{1}{6}\))
Given, 2[(\(\frac{1}{6}\) + x)(\(\frac{1}{6}\) - x) + \(\frac{1}{6}\) x \(\frac{1}{6}\) + \(\frac{1}{6}\) x \(\frac{1}{6}\)] = \(\frac{13}{96}\)
⇒ \(\frac{1}{36}\) - x2 + \(\frac{2}{36}\) = \(\frac{13}{96 \times 2}\)
⇒ \(\frac{1}{12}\) - x2 = \(\frac{13}{96 \times 2}\)
⇒ x2 = \(\frac{1}{12}\) - \(\frac{13}{96 \times 2}\) = \(\frac{1}{12}\) (1 - \(\frac{13}{16}\))
= \(\frac{1}{12}\)(\(\frac{16 -13}{16}\))
Therefore, x = \(\frac{1}{18}\)