An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colour is :

Question

Accepted Answer

To determine the probability that the three balls drawn from the urn have different colors, we need to consider the total number of ways to draw three balls and the number of favorable outcomes where each ball is of a different color.

**Step 1: Calculate the total number of ways to draw three balls from the urn.**

The urn contains a total of 9 balls (3 red, 4 blue, and 2 green). The number of ways to choose 3 balls out of 9 is given by the combination formula:

$$
\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84
$$

**Step 2: Calculate the number of favorable outcomes where the balls are of different colors.**

To have one ball of each color, we need to choose 1 red, 1 blue, and 1 green ball. The number of ways to choose 1 red ball from 3 red balls is:

$$
\binom{3}{1} = 3
$$

The number of ways to choose 1 blue ball from 4 blue balls is:

$$
\binom{4}{1} = 4
$$

The number of ways to choose 1 green ball from 2 green balls is:

$$
\binom{2}{1} = 2
$$

Thus, the number of favorable outcomes is:

$$
3 \times 4 \times 2 = 24
$$

**Step 3: Calculate the probability.**

The probability that the three balls drawn have different colors is the ratio of the number of favorable outcomes to the total number of outcomes:

$$
\frac{24}{84} = \frac{2}{7}
$$

Therefore, the probability that the three balls have different colors is $\frac{2}{7}$, which corresponds to Option 2.

**Explanation of why other options are incorrect:**

- **Option 1: $\frac{1}{3}$**

This option suggests a probability that is too high given the specific constraints of the problem. The calculation shows that the correct probability is $\frac{2}{7}$, not $\frac{1}{3}$.

- **Option 3: $\frac{1}{21}$**

This option suggests a probability that is too low. The calculation of favorable outcomes and total outcomes clearly shows that the probability is $\frac{2}{7}$.

- **Option 4: $\frac{2}{23}$**

This option is also incorrect as it does not match the calculated probability of $\frac{2}{7}$. The denominator in this option does not reflect the total number of ways to draw three balls from the urn.

In conclusion, the correct probability that the three balls drawn have different colors is $\frac{2}{7}$, which is why Option 2 is the correct answer.