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  • A bag contains 5 red marbles and 3 black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?

A bag contains 5 red marbles and 3 black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?

Answers (1)

Given

A bag contains 5 red marbles and 3 black marbles

If the first marble is red, the following conditions have to be followed for at least one marble to be black.

  1. Second marble is black and third is red = E_1
  2.  Second and third, both marbles are black = E_2

(iii) Second marble is red and third marble is black = E_3
Let event R_n = drawing red marble in n^{th} draw
And event R_n = drawing black marble in n^{th} draw

\\\therefore P(E_ 1)=P\left(R_{1}\right) \cdot P\left(B_{1} / R_{1}\right) \cdot P\left(R_{2} / B_{1} R_{1}\right)=\frac{5}{8} \times \frac{3}{7} \times \frac{4}{6}=\frac{5}{28}$ \\$\therefore \mathrm{P}\left(\mathrm{E}_{2}\right)=\mathrm{P}\left(\mathrm{R}_{1}\right) \cdot \mathrm{P}\left(\mathrm{B}_{1} / \mathrm{R}_{1}\right) \cdot \mathrm{P}\left(\mathrm{B}_{2} / \mathrm{B}_{1} \mathrm{R}_{1}\right)=\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6}=\frac{5}{56}$ \\$\therefore \mathrm{P}\left(\mathrm{E}_{3}\right)=\mathrm{P}\left(\mathrm{R}_{1}\right) \cdot \mathrm{P}\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right) \cdot \mathrm{P}\left(\mathrm{B}_{1} / \mathrm{R}_{1} \mathrm{R}_{2}\right)=\frac{5}{8} \times \frac{4}{7} \times \frac{3}{6}=\frac{5}{28}$    
Hence, the for the probability for at least one marble to be black is P

(\mathrm{E})=\mathrm{P}\left(\mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right)\\$ Therefore, $P(E)=\frac{5}{28}+\frac{5}{56}+\frac{5}{28}=\frac{25}{56}$

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