Two numbers `b\ a n d\ c` are chosen at random with replacement from the numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9. Find the probability that `x^2+b x+c >0\ for\ a l l\ x in Rdot`

Options

A. `(10)/(27)`
B. `(31)/(81)`
C. `(32)/(81)`
D. `(11)/(27)`

Answer

C

Explanation

Since b and c both can assume values from 1 to 9. So, total number of ways of choosing b and c is `9xx9=81`.
Now,
`x^(2)+bx+c gt 0 " for all " x in R`
`implies` Disc `lt 0 " i.e.," b^(2)-4c lt 0`.
The following table shows the possible values of b and c for which `b^(2)-4c lt 0`.

So, favourable number of ways =32.
Hence, required probability `=32//81`.