A bag contains 4 red and 6 black balls. Two balls are drawn in succession without replacement. The probability that the first is red and the second is black is

$\frac{4}{15}$

The correct answer is Option (3) → $\frac{4}{15}$ **

Total balls = 4 red + 6 black = 10

Probability(first red) = $\frac{4}{10}$

After drawing one red, remaining balls = 9 (3 red + 6 black)

Probability(second black) = $\frac{6}{9}$

Required probability = $\frac{4}{10} \times \frac{6}{9} = \frac{24}{90} = \frac{4}{15}$

Final Answer:

$\frac{4}{15}$