MHT CET · Maths · Probability
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
- A \(1 / 5\)
- B \(4 / 5\)
- C \(3 / 5\)
- D \(2 / 5\)
Answer & Solution
Correct Answer
(D) \(2 / 5\)
Step-by-step Solution
Detailed explanation
\(E_{1}\) can occur as \(\{(1,5),(5,1,),(2,4),(4,2),\),
\((3,3)\}\)
\(
\begin{array}{l}
E_{2} \text { -as }\{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\
(6,2),(5,2),(4,2),(3,2),(1,2)\}
\end{array}
\)
\(\therefore P\left(E_{2} / E_{1}\right)=\) Probability of \(E_{2}\) when \(E_{1}\) has occured
\(
=2 / 5
\)
\((3,3)\}\)
\(
\begin{array}{l}
E_{2} \text { -as }\{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\
(6,2),(5,2),(4,2),(3,2),(1,2)\}
\end{array}
\)
\(\therefore P\left(E_{2} / E_{1}\right)=\) Probability of \(E_{2}\) when \(E_{1}\) has occured
\(
=2 / 5
\)
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