Question

a bag contains different colored beads. the probability of drawing two black beads from the bag without replacement is \\(\frac{3}{35}\\), and the probability of drawing one black bead is \\(\frac{3}{10}\\). what is the probability of drawing a second black bead, given that the first bead is black? \\(\circ\\ \frac{1}{5}\\) \\(\circ\\ \frac{9}{350}\\) \\(\circ\\ \frac{2}{7}\\) \\(\circ\\ \frac{6}{35}\\)

Explanation:

Step1: Recall conditional probability formula

Let \( P(A) = \) probability first bead is black (\( \frac{3}{10} \)), \( P(A \cap B) = \) probability two black beads (\( \frac{3}{35} \)). Conditional probability: \( P(B|A) = \frac{P(A \cap B)}{P(A)} \)

Step2: Substitute values

\( P(B|A) = \frac{\frac{3}{35}}{\frac{3}{10}} = \frac{3}{35} \times \frac{10}{3} = \frac{10}{35} = \frac{2}{7} \)

Answer:

C. \( \frac{2}{7} \)