InterviewSolution
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InterviewSolution
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The life in hours of a ratio tube is continuous random variable with pdf `f(x)={{:((100)/x^2","xge100),(0",""else where"):}` Then, the probability that the life of tube will than 200 h if it is known that th tube is still functioning after 150 h of services isA. `1/4`B. `1/3`C. `1/2`D. None of these |
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Answer» Correct Answer - A `thereforeP[(xlt200)|(xgt150)]` `=(P(150ltxlt200))/(P(xgt150))` `(overset(200)underset(150)int(100)/x^2dx)/(oversetoounderset(150)int(100)/x^2dx)=([(-100)/x]_(150)^(200))/([(-100)/x]_(150)^oo)` `=(-[(100)/(200)-(100)/(150)])/(-[0-(100)/(150)])=(1/6)/(2/3)=1/4` |
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