Statistics And Probability Question 561

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

[Pb. CET 2004]

Options:

A) P (only one of them occurs) $ =P({{\bar{E}}_1}E_2E_3+E_1{{\bar{E}}_2}E_3+E_1E_2{{\overline{E}}_3}) $

B) P (none of them occurs) $ =P({{\overline{E}}_1}+{{\overline{E}}_2}+{{\overline{E}}_3}) $

C) P (at least 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 $ .

Show Answer

Answer:

Correct Answer: C

Solution:

  •                 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_2E_3+E_1{{\bar{E}}_2}E_3+E_1E_2{{\bar{E}}_3}) $                     \  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}) $                     \  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) $                     \  is correct.                    P (all the three occurs)                     $ =P(E_1\cap E_2\cap E_3)\ne P(E_1+E_2+E_3) $                                  \  is not correct.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_2E_3+E_1{{\bar{E}}_2}E_3+E_1E_2{{\bar{E}}_3}) $                                  \  is incorrect. 
    


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