A basket contains $4$ red, $5$ blue and $3$ green marbles. If $2$ marbles are drawn at random from the basket, what is the probability that both are red?

### Concept & Probability with Combinations When drawing multiple items without replacement, we use combinations to figure out the total number of ways to pick the group, and the number of ways to pick our desired specific group. $$\text{Probability} = \frac{\binom{\text{Favorable}}{\text{Chosen}}}{\binom{\text{Total}}{\text{Chosen}}}$$ ### Step-by-Step Solution * **Total Outcomes:** The basket contains $4$ red, $5$ blue, and $3$ green marbles. * Total marbles = $4 + 5 + 3 = 12$. * We are drawing $2$ marbles. * $n(S) = \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 6 \times 11 = 66$. * **Favorable Outcomes:** We want both marbles to be red. There are $4$ red marbles available. * $n(E) = \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 2 \times 3 = 6$. * **Calculation:** * $P(\text{Both Red}) = \frac{n(E)}{n(S)} = \frac{6}{66}$. * **Simplify:** Divide both the numerator and denominator by $6$. * $\frac{6 \div 6}{66 \div 6} = \frac{1}{11}$. ### Exam Strategy & Shortcut Instead of combinations, use sequential dependent probability. The chance the first marble is red is $\frac{4}{12}$. Since we didn't replace it, there are now $3$ red marbles left out of $11$ total. The chance the second is red is $\frac{3}{11}$. Multiply them: $\frac{4}{12} \times \frac{3}{11} = \frac{1}{3} \times \frac{3}{11} = \frac{1}{11}$. This is usually much faster to compute. ### Common Pitfall Using independent probabilities (with replacement) instead of dependent probabilities (without replacement). Calculating $\frac{4}{12} \times \frac{4}{12} = \frac{1}{9}$ is a common mistake. ### Final Answer Therefore, the correct answer is **$\frac{1}{11}$**.