A bag contains 5 red balls and 3 green balls. A ball is selected at random and not replaced. A second ball…

A bag contains 5 red balls and 3 green balls. A ball is selected at random and not replaced. A second ball is then selected. The probability of selecting one red ball and one green ball is
  1. $\frac{15}{28}$
  2. $\frac{15}{64}$
  3. $\frac{15}{56}$
  4. $\frac{15}{112}$

Solution

$\begin{aligned} & \text { Required probability }=P(R) \cdot P\left(\frac{G}{R}\right)+P(G) \cdot P\left(\frac{R}{G}\right) \\ & =\frac{5}{8} \cdot \frac{3}{7}+\frac{3}{8} \cdot \frac{5}{7} \\ & =\frac{30}{56}=\frac{15}{28}\end{aligned}$

Asked in: MHT CET 2022 (05 Aug Shift 2)

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