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if `E_(1)` and `E_(2)` are two events such that `P(E_(1))=(1)/(4),P((E_(2))/(E_(1)))=(1)/(2)` and `P=((E_(1))/(E_(2)))=(1)/(4)`A. then `E_(1)` ad `E_(2)` are independentB. `E_(1)` and `E_(2)` are exhaustiveC. `E_(2)` is twice as likely to occur as `E_(1)`D. probabilities of the events `E_(1)capE_(2),E_(1)` and `E_(2)` are in G.P. |
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Answer» Correct Answer - A::C::D `P(E_(1))=(1)/(4)` `P_((E_(2))/(E_(1)))=(P(E_(1)capE_(2)))/(P(E_(1)))=(1)/(2)` `P(E_(1)capE_(2))=(1)/(8)` `P((E_(1))/(E_(2)))=(P(E_(1)capE_(2)))/(P_(E_(2)))=(1)/(4)` `P(E_(2))=(1)/(2)` `P_(E_(1)).P_(E_(2))=P(E_(1)capE_(2))` |
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