Q268 / 270 · NDA 2012 (Phase 2) · Mathematics

An urn contains one black ball and one green ball. A second urn contains one white and one green ball. One ball is drawn at random from each urn. What is the probability of getting at least one green ball?

  • A
    $\frac{1}{2}$
  • B
    $\frac{1}{3}$
  • C
    $\frac{2}{3}$
  • Correct answer
    D
    $\frac{3}{4}$

Solution

$\begin{aligned} & \mathrm{P}(\text { at least one green ball })=1-\mathrm{P}(\text { no ball is green }) \\ & =1-\mathrm{P}(\text { i.e., ball is white and black }) \\ & =1-\frac{1}{2} \times \frac{1}{2}=1-\frac{1}{4}=\frac{3}{4}\end{aligned}$

Hence, the correct answer is option: 'D'.

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