Problem A bag has 4 red, 5 blue, and 6 green marbles (total 15). Three marbles are drawn without replacement. Find the probability all three are different colors.

Method 1: Counting combinations

• Total ways to choose any 3 marbles: C(15,3) = 455.
• Favorable ways (one of each color):
  • Choose 1 red: C(4,1) = 4
  • Choose 1 blue: C(5,1) = 5
  • Choose 1 green: C(6,1) = 6
  • Total favorable = 4 × 5 × 6 = 120

Probability = 120 / 455 = 24 / 91.

Method 2: Ordered draws (check)

• Favorable ordered sequences: pick specific marbles and order them: 4 × 5 × 6 × 3! = 720
• Total ordered sequences: 15 × 14 × 13 = 2730

Probability = 720 / 2730 = 24 / 91.

Answer 24/91

Here is the step-by-step solution to find the probability of drawing three marbles of different colours.

Total Combinations

First, we calculate the total number of ways to draw 3 marbles from the bag without replacement.

• Total marbles: 4 (red) + 5 (blue) + 6 (green) = 15 marbles

Using the combinations formula, the number of possible ways to choose 3 marbles out of 15 is: $$ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 $$

There are 455 total possible outcomes.

Successful Combinations

Next, we calculate the number of ways to successfully draw exactly one marble of each colour (1 red, 1 blue, and 1 green).

We find the combinations for each colour individually:

• Ways to choose 1 red marble out of 4: 4
• Ways to choose 1 blue marble out of 5: 5
• Ways to choose 1 green marble out of 6: 6

Multiply these together to find the total number of successful combinations: $$ 4 \times 5 \times 6 = 120 $$

There are 120 ways to draw three different colours.

Final Probability

The probability is the number of successful combinations divided by the total possible combinations: $$ \text{Probability} = \frac{120}{455} $$

To simplify the fraction, we divide the numerator and the denominator by their greatest common divisor, which is 5:

• $120 \div 5 = 24$
• $455 \div 5 = 91$

Final Answer: The probability that all three marbles are different colours is 24/91.

Probability That All Three Marbles Are Different Colours

Setting Up

The bag contains:

• 4 red
• 5 blue
• 6 green

That is 15 marbles in total, and we draw 3 without replacement. Since order doesn't matter, we can count combinations.

Step 1: Total Possible Outcomes

The number of ways to choose any 3 marbles from 15 is:

$$ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = \frac{2730}{6} = 455 $$

Step 2: Favourable Outcomes

For all three to be different colours, we need exactly one red, one blue and one green:

$$ \binom{4}{1} \times \binom{5}{1} \times \binom{6}{1} = 4 \times 5 \times 6 = 120 $$

Step 3: Compute the Probability

$$ P(\text{all different}) = \frac{120}{455} $$

Simplify by dividing numerator and denominator by 5:

$$ \frac{120}{455} = \frac{24}{91} $$

Check Using Sequential Draws

Find the probability of one particular colour order, such as Red → Blue → Green:

$$ \frac{4}{15} \times \frac{5}{14} \times \frac{6}{13} = \frac{120}{2730} $$

There are $3! = 6$ possible orders of the three colours, and each has the same probability:

$$ 6 \times \frac{120}{2730} = \frac{720}{2730} = \frac{24}{91} \checkmark $$

Answer

$$ \boxed{P = \dfrac{24}{91} \approx 0.264} $$

There is roughly a 26.4% chance that the three marbles drawn are all different colours.