MCQ
Let ${E_1},{E_2},{E_3}$ be three arbitrary events of a sample space $S$. Consider the following statements which of the following statements are correct
  • A
    $P$ (only one of them occurs)

    $ = P({\bar E_1}{E_2}{E_3} + {E_1}{\bar E_2}{E_3} + {E_1}{E_2}{\overline E _3})$

  • B
    $P$ (none of them occurs)

    $ = P({\overline E _1} + {\overline E _2} + {\overline E _3})$

  • $P$ (atleast one of them occurs)

    $ = P({E_1} + {E_2} + {E_3})$

  • D
    $P$ (all the three occurs)$ = P({E_1} + {E_2} + {E_3})$

    where $P({E_1})$denotes the probability of ${E_1}$ and ${\bar E_1}$ denotes complement of ${E_1}$.

Answer

Correct option: C.
$P$ (atleast one of them occurs)

$ = P({E_1} + {E_2} + {E_3})$

c
(c) $P$ (only one of them occurs)

$ = P({E_1}{\bar E_2}{\bar E_3} + {\bar E_1}{E_2}{\bar E_3} + {\bar E_1}{\bar E_2}{E_3})$

$ \ne P({\bar E_1}{E_2}{E_3} + {E_1}{\bar E_2}{E_3} + {E_1}{E_2}{\bar E_3})$

$\therefore$ $(a)$ is incorrect.

$P$ (none of them occurs)

$ = P({\bar E_1} \cap {\bar E_2} \cap {\bar E_3}) \ne P({\bar E_1} + {\bar E_2} + {\bar E_3})$

$\therefore$ $(b)$ is not correct.

$P$ (atleast one of them occurs)

$ = P({E_1} \cup {E_2} \cup {E_3}) = P({E_1} + {E_2} + {E_3})$

$\therefore$ $ (c) $ is correct.

$P$ (all the three occurs)

$ = P({E_1} \cap {E_2} \cap {E_3}) \ne P({E_1} + {E_2} + {E_3})$

$\therefore$ $(d)$ is not correct.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free