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  4. A bowl has 6 red marbles and 3 green marbles. The probability that a blind folded person will draw a red marble on the second draw from the bowl without replacing the marble from the first draw, is

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Q. A bowl has 6 red marbles and 3 green marbles. The probability that a blind folded person will draw a red marble on the second draw from the bowl without replacing the marble from the first draw, is

Probability - Part 2

Solution:

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$E$ : Event that the $2^{\text {nd }}$ drawn marble is red; $R: 1^{\text {st }}$ drawn is red; $G=1^{\text {st }}$ drawn is green
$P(E) =P(R \cap E)+P(G \cap E) $
$=P(R) \cdot P(E / R)+P(G) \cdot P(E / G) $
$=\frac{6}{9} \cdot \frac{5}{8}+\frac{3}{9} \cdot \frac{6}{8}=\frac{48}{72}=\frac{2}{3} $

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