Given set : {1, 2, 3, 4 ..., 14}
⇒ Total no. of outcomes = 14
Given :
f(x)=(x–2)(x2–9)(x–6)(x–8)⇒f(x)=(x–2)(x+3)(x–3)(x–6)(x–8)
Its solutions are x = 2, –3, 3, 6 and 8.
But, of these solutions only 2 and 3 prime number lying in the given set.
∴ No. of favourable outcomes = 2
∴ Required probability
=