Q265 / 270 · NDA 2015 (Phase 2) · Mathematics

A bag contains 4 white and 2 black balls and another bag contains 3 white and 5 black balls. If one ball is drawn from each bag, then the probability that one ball is white and one ball is black is

  • A
    5/24
  • Correct answer
    B
    13/24
  • C
    1/4
  • D
    2/3

Solution

$\begin{aligned} & \mathrm{P} \text { (one white ball \& one black ball) } \\ & =\mathrm{P}\{[\text { black from 1st bag \& white from 2nd] or [white } \\ & \text { from 1st \& black from 2nd] \} } \\ & =\mathrm{P}\left(\left[\mathrm{~B}_1 \cap \mathrm{~W}_2\right) \cup\left(\mathrm{W}_1 \cap \mathrm{~B}_2\right)\right] \\ & =\mathrm{P}\left(\mathrm{~B}_1 \cap \mathrm{~W}_2\right)+\mathrm{P}\left(\mathrm{~W}_1 \cap \mathrm{~B}_2\right) \\ & \text { (By addition theorem for } \\ & \text { mutually exclusive events) } \\ & =\mathrm{P}\left(\mathrm{~B}_1\right) \mathrm{P}\left(\mathrm{~W}_2\right)+\mathrm{P}\left(\mathrm{~W}_1\right) \mathrm{P}\left(\mathrm{~B}_2\right) \\ & \left(\therefore \mathrm{B}_2 \& \mathrm{~W}_2 ; \mathrm{B}_2 \& \mathrm{~W}_1\right. \text { are pairs } \\ & \text { of independent events) } \\ & =\left[\frac{2}{6} \times \frac{3}{8}\right]+\left[\frac{4}{6} \times \frac{5}{8}\right] \\ & =\frac{13}{24} \end{aligned}$

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

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