CUET Preparation Today
A bag contains 5 red and 7 black balls. Three balls are chosen at random one by one without replacement. What is the probability that the third ball is a Red ball given that first ball is red and second ball is black? |
$\frac{7}{33}$ $\frac{14}{33}$ $\frac{7}{66}$ $\frac{2}{5}$ |
$\frac{2}{5}$ |
The correct answer is Option 4: $\frac{2}{5}$ Initial: 5 Red, 7 Black (Total 12) 1st Draw (Red): 1 Red is removed $\rightarrow$ 4 Red left. 2nd Draw (Black): 1 Black is removed $\rightarrow$ 6 Black left. Now, there are 10 balls left in the bag ($4 \text{ Red} + 6 \text{ Black}$). The probability of picking a Red ball on the third draw is:
$P = \frac{\text{Red balls remaining}}{\text{Total balls remaining}} = \frac{4}{10} = \frac{2}{5}$
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