Question

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) $= P \left( \bar { E } _ { 1 } E _ { 2 } E _ { 3 } + E _ { 1 } \bar { E } _ { 2 } E _ { 3 } + E _ { 1 } E _ { 2 } \bar { E } _ { 3 } \right)$
• B. P (none of them occurs) $= P \left( \bar { E } _ { 1 } + \bar { E } _ { 2 } + \bar { E } _ { 3 } \right)$
• C. P (atleast one of them occurs) $= P \left( E _ { 1 } + E _ { 2 } + E _ { 3 } \right)$
• D. P (all the three occurs)$= P \left( E _ { 1 } + E _ { 2 } + E _ { 3 } \right)$ Where $P \left( E _ { 1 } \right)$ denotes the probability of $E _ { 1 }$ and $\bar { E } _ { 1 }$ denotes complement of$E _ { 1 }$.=$P \left( E _ { 1 } + E _ { 2 } + E _ { 3 } \right)$

Answer 1

Answer: (C)

P (atleast one of them occurs) $= P \left( E _ { 1 } + E _ { 2 } + E _ { 3 } \right)$

Answer 2

P (only one of them occurs)

$= P \left( E _ { 1 } \bar { E } _ { 2 } \bar { E } _ { 3 } + \bar { E } _ { 1 } E _ { 2 } \bar { E } _ { 3 } + \bar { E } _ { 1 } \bar { E } _ { 2 } E _ { 3 } \right)$

∴ (1) is incorrect.

P (none of them occurs)

$= P \left( \bar { E } _ { 1 } \cap \bar { E } _ { 2 } \cap \bar { E } _ { 3 } \right) \neq P \left( \bar { E } _ { 1 } + \bar { E } _ { 2 } + \bar { E } _ { 3 } \right)$

∴ (2) is not correct.

P (atleast one of them occurs)

$= P \left( E _ { 1 } \cup E _ { 2 } \cup E _ { 3 } \right) = P \left( E _ { 1 } + E _ { 2 } + E _ { 3 } \right)$

∴ (3) is correct.

P (all the three occurs)

$= P \left( E _ { 1 } \cap E _ { 2 } \cap E _ { 3 } \right) \neq P \left( E _ { 1 } + E _ { 2 } + E _ { 3 } \right)$

∴ (4) is incorrect