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A die is rolled twice and the sum of the numbers appearing on them is observed to be 7 . What is the conditional probability that the number 2 has appeared at least once?
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The correct answer is:
$\frac{1}{3}$
Let $\mathrm{A}$ and $\mathrm{B}$ be two events such that $\mathrm{A}=$ getting number 2 at least once
$B=$ getting 7 as the sum of the numbers on two dice
Here, $\mathrm{A}=\{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$ $(1,2),(3,2),(4,2),(5,2),(6,2)\}$ and $\mathrm{B}=\{(2,5),(5,2),(6,1),(1,6),(3,4),(4,3)\}$ $\therefore \mathrm{P}(\mathrm{A})=\frac{11}{36}, \mathrm{P}(\mathrm{B})=\frac{6}{36}$ $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{36}$ $\therefore$ Required probability $\mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{2 / 36}{6 / 36}=\frac{2}{6}=\frac{1}{3}$
$B=$ getting 7 as the sum of the numbers on two dice
Here, $\mathrm{A}=\{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$ $(1,2),(3,2),(4,2),(5,2),(6,2)\}$ and $\mathrm{B}=\{(2,5),(5,2),(6,1),(1,6),(3,4),(4,3)\}$ $\therefore \mathrm{P}(\mathrm{A})=\frac{11}{36}, \mathrm{P}(\mathrm{B})=\frac{6}{36}$ $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{36}$ $\therefore$ Required probability $\mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{2 / 36}{6 / 36}=\frac{2}{6}=\frac{1}{3}$
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