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?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\frac{2}{5}\)
Since the first two draws are known (one red, then one black, both not replaced), the bag now holds
\( 5 - 1 = 4 \) red balls
\( 7 - 1 = 6 \) black balls
\( 4 + 6 = 10 \) balls in all
The conditional probability of a red ball on the third draw depends only on these 10 balls:
\( \displaystyle P = \frac{4}{10} = \frac{2}{5} \)
Tip: “Given that” is not the same as “and”. The probability of red, black, red in that order is \(\frac{5}{12} \times \frac{7}{11} \times \frac{4}{10} = \frac{7}{66}\), which is Option (3), not the probability asked for.