A bag contains $3$ red, $5$ blue, and $2$ green balls. One ball is drawn at random. What is the probability that the drawn ball is **not** red?
A$\dfrac{3}{10}$
B$\dfrac{7}{10}$
C$\dfrac{1}{2}$
D$\dfrac{2}{5}$
Answer & Solution
Correct answer: B. $\dfrac{7}{10}$
Total balls: $3 + 5 + 2 = 10$. Non-red balls: $5 + 2 = 7$. So $P(\text{not red}) = \dfrac{7}{10}$.
Alternative path using the complement rule, which is often faster: $P(\text{not red}) = 1 - P(\text{red}) = 1 - \dfrac{3}{10} = \dfrac{7}{10}$. Either approach gives the same answer; the complement rule generalises nicely when there are many alternative colours.
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