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If $E_{1}$ denotes the events of coming sum 6 in throwing two dice and $E_{2}$ be the event of coming 2 in any one of the two, then $P\left(E_{2} / \mathrm{E}_{1}\right.$ ) is
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Verified Answer
The correct answer is:
$2 / 5$
$E_{1}$ can occur as $\{(1,5),(5,1,),(2,4),(4,2),$,
$(3,3)\}$
$$
\begin{array}{l}
E_{2} \text { -as }\{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\
(6,2),(5,2),(4,2),(3,2),(1,2)\}
\end{array}
$$
$\therefore \quad P\left(E_{2} / E_{1}\right)=$ Probability of $E_{2}$ when $E_{1}$ has occured
$$
=2 / 5
$$
$(3,3)\}$
$$
\begin{array}{l}
E_{2} \text { -as }\{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\
(6,2),(5,2),(4,2),(3,2),(1,2)\}
\end{array}
$$
$\therefore \quad P\left(E_{2} / E_{1}\right)=$ Probability of $E_{2}$ when $E_{1}$ has occured
$$
=2 / 5
$$
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