InterviewSolution
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InterviewSolution
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Let `Ea n dF`be tow independent events. The probability that exactly one of themoccurs is 11/25 and the probability if none of them occurring is 2/25. If `P(T)`deontes the probability of occurrence of the event `T ,`then`P(E)=4/5,P(F)=3/5``P(E)=1/5,P(F)=2/5``P(E)=2/5,P(F)=1/5``P(E)=3/5,P(F)=4/5`A. `P(E )=(4)/(5),P(F)=(3)/(5)`B. `P(E )=(1)/(5),P(F)=(2)/(5)`C. `P(E )=(2)/(5),P(F)=(1)/(5)`D. `P(E )=(6)/(5),P(F)=(1)/(5)` |
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Answer» Correct Answer - A We have, `P(E )+P(F)-2P(E cap F)=(11)/(25) " and" P(overline(E ) cap overline(F))=(2)/(25)` `implies P(E )+P(F)-2P(E )P(F)=(11)/(25) " and "P(overline(E ))P(overline(F))=(2)/(25)` `implies x+y-2xy=(11)/(25) " and "1-x+xy=(2)/(25)`, where P(E ) =x and P(F)=y `implies x+y+2-2x-2y=(11)/(25)+2xx(2)/(25)` [ On eliminating xy] `implies x+y=(7)/(5) implies y=(7)/(5) -x` Substituting `y=(7)/(5)-x " in " 1-x-y+xy=(2)/(25)`, we get `1-(7)/(5)+x((7)/(5)-x)=(2)/(25)` `implies 25x^(2)-35x+12=0 implies x=(3)/(5), (4)/(5)` When `x=(3)/(5), y=(4)/(5) " and " y=(3)/(5) " for " x=(4)/(5)` Hence, `P(E )=(3)/(5), P(F)=(4)/(5) " or " P(E )=(4)/(5), P(F)=(3)/(5)` |
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