This problem involves calculating the probability of selecting a specific combination of balls from a box when balls are drawn without replacement. We have a total of 10 balls, consisting of 4 red balls and 6 black balls.

Balls in the Box

First, let's identify the total number of balls and the count of each color:

• Total number of red balls (R) = 4
• Total number of black balls (B) = 6
• Total number of balls in the box = 4 + 6 = 10 balls

We are selecting three balls randomly from the box, one after another, without replacement. This means that once a ball is drawn, it is not put back into the box, so the total number of balls available decreases with each subsequent draw.

Desired Selection Probability

We want to find the probability that the selected set contains exactly one red ball and two black balls.

Method 1: Using Combinations for Ball Selection

Since the question asks for the composition of the "selected set" (one red ball and two black balls), the order in which the balls are drawn does not matter for the final set. Therefore, we can use combinations.

Total Possible Outcomes

The total number of ways to select 3 balls from the 10 available balls is given by the combination formula:

\(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

Here, \(n = 10\) (total balls) and \(k = 3\) (balls to be selected).

\(\text{Total ways to select 3 balls} = \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1}\)

\(\text{Total ways to select 3 balls} = 10 \times 3 \times 4 = 120\)

Favorable Outcomes for Ball Selection

We need to select 1 red ball from the 4 red balls AND 2 black balls from the 6 black balls.

• Ways to select 1 red ball from 4 = \(\binom{4}{1} = 4\)
• Ways to select 2 black balls from 6 = \(\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15\)

The total number of favorable ways to select 1 red ball and 2 black balls is the product of these two combinations:

\(\text{Favorable ways} = \binom{4}{1} \times \binom{6}{2} = 4 \times 15 = 60\)

Calculating Probability for Selected Balls

The probability is the ratio of favorable outcomes to the total possible outcomes:

\(\text{Probability (1 Red, 2 Black)} = \frac{\text{Favorable outcomes}}{\text{Total possible outcomes}} = \frac{60}{120}\)

\(\text{Probability (1 Red, 2 Black)} = \frac{1}{2}\)

Method 2: Sequential Probability (Considering Order of Ball Selection)

Even though the final set's order doesn't matter, we can also consider the sequential probability, which accounts for the specific order of drawing the balls and then sums up the probabilities of all valid sequences.

The possible sequences to get one red ball and two black balls are:

• Red, Black, Black (RBB)
• Black, Red, Black (BRB)
• Black, Black, Red (BBR)

Probability of RBB Sequence

• Probability of drawing a Red ball first = \(\frac{4}{10}\)
• Probability of drawing a Black ball second (given a Red was drawn) = \(\frac{6}{9}\)
• Probability of drawing a Black ball third (given R, B were drawn) = \(\frac{5}{8}\)

\(P(\text{RBB}) = \frac{4}{10} \times \frac{6}{9} \times \frac{5}{8} = \frac{120}{720} = \frac{1}{6}\)

Probability of BRB Sequence

• Probability of drawing a Black ball first = \(\frac{6}{10}\)
• Probability of drawing a Red ball second (given a Black was drawn) = \(\frac{4}{9}\)
• Probability of drawing a Black ball third (given B, R were drawn) = \(\frac{5}{8}\)

\(P(\text{BRB}) = \frac{6}{10} \times \frac{4}{9} \times \frac{5}{8} = \frac{120}{720} = \frac{1}{6}\)

Probability of BBR Sequence

• Probability of drawing a Black ball first = \(\frac{6}{10}\)
• Probability of drawing a Black ball second (given a Black was drawn) = \(\frac{5}{9}\)
• Probability of drawing a Red ball third (given B, B were drawn) = \(\frac{4}{8}\)

\(P(\text{BBR}) = \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} = \frac{120}{720} = \frac{1}{6}\)

Total Probability from Sequential Draws

The total probability of getting one red ball and two black balls is the sum of the probabilities of these mutually exclusive sequences:

\(P(\text{1 Red, 2 Black}) = P(\text{RBB}) + P(\text{BRB}) + P(\text{BBR})\)

\(P(\text{1 Red, 2 Black}) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\)

Conclusion on Ball Probability

Both methods yield the same result. The probability that the selected set contains one red ball and two black balls when three balls are selected randomly without replacement from a box of 4 red and 6 black balls is \(\frac{1}{2}\).