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 be less than 200 h if it is known that the tube is still functioning after 150 h of services is

Options

A. `1/4`
B. `1/3`
C. `1/2`
D. None of these

Answer

A

Explanation

`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`