A bag has $3$ red, $4$ blue and $5$ green balls. Probability that a randomly drawn ball is NOT green is:

A bag has $3$ red, $4$ blue and $5$ green balls. Probability that a randomly drawn ball is NOT green is:
  1. $\dfrac{5}{12}$
  2. $\dfrac{7}{12}$
  3. $\dfrac{1}{2}$
  4. $\dfrac{1}{3}$

Solution

Total $= 12$. Not green $= 3+4 = 7$. $P = \dfrac{7}{12}$.

Asked in: IMO

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