Step 1: Set up the events.
Let $B$ be the event that an individual is blue, so $P(B) = 0.4$. Let $M$ be the event that an individual is male, so $P(M) = 0.3$. We want the probability that both $B$ and $M$ happen for the same individual, written $P(B \cap M)$.
Step 2: Apply the independence rule.
The question states colour and sex are independent of each other. For two independent events, knowing one event happened tells you nothing about whether the other happened, and the joint probability is found simply by multiplying the two separate probabilities together:
\[ P(B \cap M) = P(B) \cdot P(M) \]
This is different from events that are dependent, where you would instead need a conditional probability like $P(M \mid B)$, which is not required here since independence is explicitly given.
Step 3: Substitute and compute.
\[ P(B \cap M) = 0.4 \times 0.3 = 0.12 \]
We can also sanity check this using a simple table: imagine 1000 birds. 400 of them are blue (0.4 x 1000). Among any subgroup, 30 percent are expected to be male, since sex is independent of colour, so 30 percent of those 400 blue birds, which is 120 birds, are blue males. That gives 120 out of 1000, or 0.12, matching the direct multiplication.
Step 4: Final answer.
Rounded to two decimal places, the probability of picking a blue male at random is 0.12.
\[ \boxed{0.12} \]