Conditional probability

Suppose a batch contains ten items, of which four are defective. Two items are drawn at random from the batch, one after the other, without replacement. What is the probability that:
I) both are defective?
Ii) Is the second item defective?

Correct answer:

p₁ =  0.1333
p₂ =  0.4

Step-by-step explanation:

$$\begin{gathered} r=\frac{4}{10}=\frac{2}{5}=0.4 \\ {p}_1=\frac{4}{10}\cdot \frac{3}{9}=\frac{4\cdot 3}{10\cdot 9}=\frac{12}{90}=\frac{2}{15}=0.1333 \end{gathered}$$

$p_2=\frac{4}{10}\cdot \frac{3}{9}+\frac{6}{10}\cdot \frac{4}{9}=\frac{2}{5}=0.4$

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