Introduction:
The problem involves drawing 3 balls from an urn containing 6 white and 4 black balls. We need to find the expected number of black balls that will be obtained.

Method:
To solve the problem, we need to use the concept of probability. Let us denote the probability of drawing a black ball by P(B) and the probability of drawing a white ball by P(W).

Step 1: Find the probability of drawing 1 black ball
The probability of drawing a black ball on the first draw is 4/10. After the first ball is drawn, there are 3 black balls left in the urn and 9 total balls. Therefore, the probability of drawing a black ball on the second draw, given that a black ball was not drawn on the first draw, is 3/9. Similarly, the probability of drawing a black ball on the third draw, given that black balls were not drawn on the first two draws, is 2/8.
Therefore, the probability of drawing exactly 1 black ball is:
P(1B) = P(B) x P(W) x P(W) + P(W) x P(B) x P(W) + P(W) x P(W) x P(B)
P(1B) = 4/10 x 6/9 x 5/8 + 6/10 x 4/9 x 5/8 + 6/10 x 5/9 x 4/8 = 15/32

Step 2: Find the probability of drawing 2 black balls
The probability of drawing exactly 2 black balls is:
P(2B) = P(B) x P(B) x P(W) + P(B) x P(W) x P(B) + P(W) x P(B) x P(B)
P(2B) = 4/10 x 3/9 x 6/8 + 4/10 x 6/9 x 3/8 + 6/10 x 4/9 x 3/8 = 27/80

Step 3: Find the probability of drawing 3 black balls
The probability of drawing exactly 3 black balls is:
P(3B) = P(B) x P(B) x P(B) = 4/10 x 3/9 x 2/8 = 1/60

Step 4: Find the expected number of black balls
The expected number of black balls is the sum of the product of the number of black balls and their respective probabilities:
E(X) = 1 x P(1B) + 2 x P(2B) + 3 x P(3B)
E(X) = 1 x 15/32 + 2 x 27/80 + 3 x 1/60 = 0.9

Conclusion:
Therefore, the expected number of black balls that will be obtained is 0.9. This means that on average, we can expect to obtain less than one black ball.