A bag contains 5 red and 4 green balls. Four balls are drawn at random then find the probability that two balls are of red colour and two balls are of green.

Answer

To solve the problem of finding the probability of drawing 2 red balls and 2 green balls from a bag containing 5 red and 4 green balls, we can follow these steps:

### Step 1: Determine the total number of ways to draw 4 balls from the bag.

The total number of balls in the bag is:
- Red balls = 5
- Green balls = 4
- Total balls = 5 + 4 = 9

The total number of ways to choose 4 balls from 9 is given by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \).

So, we calculate:
\[ C(9, 4) = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \]

### Step 2: Determine the number of ways to draw 2 red balls and 2 green balls.

We need to find:
1. The number of ways to choose 2 red balls from 5.
2. The number of ways to choose 2 green balls from 4.

Calculating these:
- For red balls:
\[ C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10 \]

- For green balls:
\[ C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6 \]

### Step 3: Calculate the total number of favorable outcomes.

The total number of favorable outcomes (drawing 2 red and 2 green) is the product of the combinations calculated in Step 2:
\[ \text{Favorable outcomes} = C(5, 2) \times C(4, 2) = 10 \times 6 = 60 \]

### Step 4: Calculate the probability.

The probability \( P \) of drawing 2 red and 2 green balls is given by the ratio of the number of favorable outcomes to the total number of outcomes:
\[ P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{60}{126} \]

### Step 5: Simplify the probability.

Now we simplify \( \frac{60}{126} \):
\[ P = \frac{60 \div 6}{126 \div 6} = \frac{10}{21} \]

### Final Answer:
The probability that 2 balls are of red color and 2 balls are of green color is \( \frac{10}{21} \).

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