An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If two marbles are picked up at random, what is the probability that either both are green or both are yellow?

### Concept & The "OR" Rule in Probability When a question asks for the probability of Event A "OR" Event B (mutually exclusive events), you ADD their individual numbers of favorable outcomes. $$P(A \text{ or } B) = \frac{^{n_A}C_r + ^{n_B}C_r}{^NC_r}$$ ### Step-by-Step Solution 1. **Total Items:** $6$ red, $4$ blue, $2$ green, and $3$ yellow marbles. Total $= 15$ marbles. 2. **Total Outcomes:** Drawing $2$ marbles from $15$. $$^{15}C_2 = \frac{15 \times 14}{2 \times 1} = 105$$ 3. **Favorable Outcomes:** We want either $2$ green OR $2$ yellow. - Ways to pick $2$ green out of $2 = ^2C_2 = 1$ - Ways to pick $2$ yellow out of $3 = ^3C_2 = 3$ - Total favorable ways (OR means ADD) $= 1 + 3 = 4$ 4. **Probability:** $$P(\text{both green or both yellow}) = \frac{4}{105}$$ ### Exam Strategy & Shortcut Whenever you see "$^NC_N$", it is always $1$. Similarly, $^NC_{N-1}$ is always $N$. Recognizing this saves time calculating $^2C_2$ and $^3C_2$. ### Common Pitfall Multiplying the outcomes ($1 \times 3$) instead of adding them. Remember that "OR" signifies different possible scenarios that increase your total chances, meaning addition. ### Final Answer Therefore, the correct answer is **$\frac{4}{105}$**.