Probability of Drawing Exactly Two Same Colour Balls

This problem requires us to calculate the probability of a specific outcome when selecting items randomly from a set. We have a bag with balls of three different colors (red, white, and green). We need to find the probability that out of the three balls drawn randomly, precisely two balls are of the same color.

Bag Contents and Draw Details

Let's list the information given:

• Number of red balls: 3
• Number of white balls: 4
• Number of green balls: 5
• Total number of balls in the bag: $3 + 4 + 5 = 12$
• Number of balls drawn at random: 3

Total Ways to Draw 3 Balls

The first step in finding probability is determining the total number of possible outcomes. Since the order in which the balls are drawn does not matter, we use combinations. The total number of ways to choose 3 balls from the 12 available balls is calculated using the combination formula $ \binom{n}{k} = \frac{n!}{k!(n-k)!} $, where $n$ is the total number of items and $k$ is the number of items to choose.

Total possible combinations of drawing 3 balls from 12 is:

$ \binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220 $

So, there are 220 different ways to draw 3 balls from the bag.

Ways to Draw Exactly Two Same Colour Balls

Next, we need to figure out the number of ways to draw 3 balls such that exactly two of them are the same color. This can happen in three distinct scenarios:

• Scenario 1: Picking 2 red balls and 1 ball of a different color.
• Scenario 2: Picking 2 white balls and 1 ball of a different color.
• Scenario 3: Picking 2 green balls and 1 ball of a different color.

Calculating Ways for Each Colour Combination

Let's calculate the number of ways for each scenario:

Scenario 1: Exactly 2 Red Balls

We choose 2 red balls out of 3, and 1 ball from the remaining $4 + 5 = 9$ non-red balls.

• Ways to choose 2 red balls from 3: $ \binom{3}{2} = 3 $
• Ways to choose 1 non-red ball from 9: $ \binom{9}{1} = 9 $
• Total ways for this scenario: $ \binom{3}{2} \times \binom{9}{1} = 3 \times 9 = 27 $

Scenario 2: Exactly 2 White Balls

We choose 2 white balls out of 4, and 1 ball from the remaining $3 + 5 = 8$ non-white balls.

• Ways to choose 2 white balls from 4: $ \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 $
• Ways to choose 1 non-white ball from 8: $ \binom{8}{1} = 8 $
• Total ways for this scenario: $ \binom{4}{2} \times \binom{8}{1} = 6 \times 8 = 48 $

Scenario 3: Exactly 2 Green Balls

We choose 2 green balls out of 5, and 1 ball from the remaining $3 + 4 = 7$ non-green balls.

• Ways to choose 2 green balls from 5: $ \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 $
• Ways to choose 1 non-green ball from 7: $ \binom{7}{1} = 7 $
• Total ways for this scenario: $ \binom{5}{2} \times \binom{7}{1} = 10 \times 7 = 70 $

Summing Favorable Ways

To get the total number of ways to draw exactly two balls of the same color, we sum the ways from all three scenarios:

$ \text{Total favorable ways} = 27 + 48 + 70 = 145 $

Calculating the Final Probability

Probability is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes.

$ \text{Probability} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}} $

Plugging in our numbers:

$ \text{Probability} = \frac{145}{220} $

Simplifying the Probability Fraction

The fraction $ \frac{145}{220} $ can be simplified. We can see that both the numerator (145) and the denominator (220) are divisible by 5.

$ \frac{145 \div 5}{220 \div 5} = \frac{29}{44} $

The simplified probability is $ \frac{29}{44} $.

Summary of the Probability Calculation

We calculated the total number of ways to draw 3 balls from 12, which is 220. We then identified and calculated the number of ways to draw exactly two balls of the same color, summing up to 145 favorable outcomes. The probability is the ratio of these two numbers, $ \frac{145}{220} $, which simplifies to $ \frac{29}{44} $. This represents the likelihood of the specific event occurring.