Question

A bag contains 5 red and 4 black balls. Three balls are drawn at random from the bag. The probability that two of them are red and one is black is:

• A. $\frac{5}{21}$
• B. $\frac{10}{21}$
• C. $\frac{5}{14}$
• D. $\frac{25}{84}$

Hint:

Always simplify combinations first: $^{5}C_2 = 10$ and $^{9}C_3 = 84$. This makes dividing $\frac{40}{84}$ to get $\frac{10}{21}$ straightforward.

Answer 1

The Correct Option is B

Step 1: Basic probability idea.
Probability is favourable ways over total ways. We count using combinations because order does not matter when drawing balls.

Step 2: Total balls and draws.
There are $5 + 4 = 9$ balls and we draw 3.

Step 3: Count all ways.
\[ {}^{9}C_3 = \frac{9 \times 8 \times 7}{6} = 84 \]

Step 4: Count the good ways.
Pick 2 red from 5 and 1 black from 4: \[ {}^{5}C_2 \times {}^{4}C_1 = 10 \times 4 = 40 \]

Step 5: Form the probability.
\[ P = \frac{40}{84} \]

Step 6: Simplify.
Divide top and bottom by 4: \[ P = \frac{10}{21} \] \[ \boxed{ \frac{10}{21} } \]