Probability of obtaining total sum 7 = probability of getting opposite faces.

Probability of getting opposite faces

$$ = 2\left[ {\left( {{1 \over 6} - x} \right)\left( {{1 \over 6} + x} \right) + {1 \over 6} \times {1 \over 6} + {1 \over 6} \times {1 \over 6}} \right]$$

$$ \Rightarrow 2\left[ {\left( {{1 \over 6} - x} \right)\left( {{1 \over 6} + x} \right) + {1 \over 6} \times {1 \over 6} + {1 \over 6} \times {1 \over 6}} \right] = {{13} \over {96}}$$ (given)

$$ \Rightarrow $$ $$x = {1 \over 8}$$

$$D \ne 0 \Rightarrow \left| {\matrix{ 1 & 1 & 1 \cr 1 & 2 & 3 \cr 1 & 3 & \lambda \cr } } \right| \ne 0 \Rightarrow \lambda \ne 5$$

For no solution D = 0 $$\Rightarrow$$ $$\lambda$$ = 5

$${D_1} = \left| {\matrix{ 1 & 1 & 5 \cr 1 & 2 & \mu \cr 1 & 3 & 1 \cr } } \right| \ne 0 \Rightarrow \mu \ne 3$$

$$p = {5 \over 6}$$

$$q = {1 \over 6} \times {5 \over 6} = {5 \over {36}}$$

Option (b).

P(x $$\ge$$ 5 | x > 2) = $${{P(x \ge 5)} \over {P(x > 2)}}$$

= $${{{{\left( {{5 \over 6}} \right)}^4}.{1 \over 6} + {{\left( {{5 \over 6}} \right)}^5}.{1 \over 6} + ....... + \infty } \over {{{\left( {{5 \over 6}} \right)}^2}.{1 \over 6} + {{\left( {{5 \over 6}} \right)}^3}.{1 \over 6} + ...... + \infty }}$$

=$${{{{{{\left( {{5 \over 6}} \right)}^4}.{1 \over 5}} \over {1 - {5 \over 6}}}} \over {{{{{\left( {{5 \over 6}} \right)}^2}.{1 \over 6}} \over {1 - {5 \over 6}}}}} = {\left( {{5 \over 6}} \right)^2} = {{25} \over {36}}$$

P (Exactly one of A or B)

$$ = P\left( {A \cap \overline B } \right) + \left( {\overline A \cap B} \right) = {5 \over 9}$$

$$ = P(A)P(\overline B ) + P(\overline A )P(B) = {5 \over 9}$$

$$ \Rightarrow P(A)(1 - P(B)) + (1 - P(A))P(B) = {5 \over 9}$$

$$ \Rightarrow p(1 - 2p) + (1 - p)2p = {5 \over 9}$$

$$ \Rightarrow 36{p^2} - 27p + 5 = 0$$

$$ \Rightarrow p = {1 \over 3}$$ or $${5 \over {12}}$$

$${p_{\max }} = {5 \over {12}}$$