Question

a box contains 5 black and 2 white balls. 2 balls are drawn without replacement. find the probability of drawing a black and a white ball.

Explanation:

Step1: Calculate total number of balls

Total balls = 5 (black) + 2 (white) = 7.

Step2: Calculate probability of black then white

Probability of first black: $\frac{5}{7}$. Then, after removing one black, total balls = 6, white balls = 2. Probability of white: $\frac{2}{6}$. So this path: $\frac{5}{7} \times \frac{2}{6} = \frac{10}{42}$.

Step3: Calculate probability of white then black

Probability of first white: $\frac{2}{7}$. Then, after removing one white, total balls = 6, black balls = 5. Probability of black: $\frac{5}{6}$. So this path: $\frac{2}{7} \times \frac{5}{6} = \frac{10}{42}$.

Step4: Sum the two probabilities

Total probability = $\frac{10}{42} + \frac{10}{42} = \frac{20}{42} = \frac{10}{21}$.

Answer:

$\frac{10}{21}$ (corresponding to the option with $\boldsymbol{\frac{10}{21}}$)