$\begin {array} {1 1} (1)\;\large\frac{1}{2} & \quad (2)\;\large\frac{1}{6} \\ (3)\;\large\frac{7}{12} & \quad (4)\;\large\frac{5}{12} \end {array}$

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If B throws 1 then A can throw only 1.

If B throws 2 then A can throw only 1 and 2

$ \therefore $ Required probability = $ \large\frac{1}{6}.\large\frac{1}{6}+\large\frac{1}{6}.\large\frac{2}{6}+\large\frac{1}{6}.\large\frac{3}{6}+\large\frac{1}{6}.\large\frac{4}{6}+\large\frac{1}{6}.\large\frac{5}{6}+\large\frac{1}{6}.\large\frac{6}{6}$

$ \large\frac{1}{36} [ 1+2+3+4+5+6]$

$ \large\frac{1}{36} \times 21$

$ = \large\frac{7}{12}$

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