Have no idea how to solve this task about probabilistic problem, just wonder what method should be applied

Question

task: From the box containing 7 black balls and 3 red we select two.

What is the probability of getting two black balls if we know that one of the chosen balls is black ?

Answer 1

Assuming you do not put back any of the chosen balls and that 10 is the total number of all balls in the box (thus there are no other colors in the box), then precede as follows:

first draw: P(drawing balck ball) = $\frac{7}{10}$ because you have 10 balls in total and seven of them are black. Now, since you removed one ball, there are only 9 balls in the box left. Moreover, we know that the removed ball was black. Thus, there are now 6 black balls and three red balls in the box. Thence,

second draw: P(drawing black ball) = $\frac{6}{9}$.

Now, if you read your question carefully, you notice that the problem states a conditional probability. Notice the difference between these two statements:

1. What is the probability of drawing two balck balls?
2. What is the probability of drawing two black balls, knwoing that one of the balls is black.

The formula for conditional probability is given by:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

Thus, P(second black ball, given we already have a black ball) = $\frac{\frac{7}{10} \cdot \frac{6}{9} }{\frac{7}{10}} = \frac{6}{9}.$