Question

An urn contains 9 balls of which 3 are red, 4 are blue and 2 are green. Three balls are drawn at random from the urn. The probability that the three balls have different colors is

• A. $\frac{1}{14}$
• B. $\frac{3}{14}$
• C. $\frac{1}{7}$
• D. $\frac{2}{7}$

Hint:

To minimize calculation mistakes when simplifying fractions, write out the terms explicitly without multiplying them into large values first:
$$P(E) = \frac{3 \times 4 \times 2}{\frac{9 \times 8 \times 7}{3 \times 2 \times 1}} = \frac{3 \times 4 \times 2}{3 \times 4 \times 7} = \frac{2}{7}$$ Notice how the factor $3 \times 4$ cancels out instantly!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem presents an urn filled with a total of 9 multi-colored balls. We need to calculate the probability that drawing three balls completely at random yields exactly one ball of each individual color.

Step 2: Key Formula or Approach:
The classic probability formula is given by:
$$P(E) = \frac{n(E)}{n(S)}$$ Where:
$n(S) = $ Total combinations for picking any 3 random balls out of 9 total items = $\binom{9}{3}$.
$n(E) = $ Favorable combinations for getting 1 Red, 1 Blue, and 1 Green ball = $\binom{3}{1} \times \binom{4}{1} \times \binom{2}{1}$.

Step 3: Detailed Explanation:
Let's compute the total sample space size $n(S)$:
$$n(S) = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$$ Next, let's compute the number of favorable event combinations $n(E)$:
$$n(E) = \binom{3}{1} \times \binom{4}{1} \times \binom{2}{1} = 3 \times 4 \times 2 = 24$$ Now, calculate the probability ratio $P(E)$:
$$P(E) = \frac{24}{84}$$ Divide both the numerator and the denominator by their greatest common divisor, 12:
$$P(E) = \frac{2}{7}$$

Step 4: Final Answer:
The probability of picking three balls of entirely distinct colors is $\frac{2}{7}$, matching option (D).