An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 is green, 2 are blue and 1 is red ?

### Concept & Complex Combinations For drawing multiple items of various different colors simultaneously, multiply the combinations for each specific color requirement. $$P = \frac{^AC_a \times ^BC_b \times ^CC_c}{^{\text{Total}}C_{\text{Total Drawn}}}$$ ### Step-by-Step Solution 1. **Total Items:** $6$ red, $4$ blue, $2$ green, and $3$ yellow marbles. Total $= 15$ marbles. 2. **Total Outcomes:** Drawing $4$ marbles from $15$. $$^{15}C_4 = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 15 \times 7 \times 13 = 1365$$ 3. **Favorable Outcomes:** We need $1$ green, $2$ blue, and $1$ red. - Ways to pick $1$ green from $2 = ^2C_1 = 2$ - Ways to pick $2$ blue from $4 = ^4C_2 = 6$ - Ways to pick $1$ red from $6 = ^6C_1 = 6$ - Total favorable ways $= 2 \times 6 \times 6 = 72$ 4. **Probability:** $$P = \frac{72}{1365}$$ 5. **Simplify:** Both $72$ and $1365$ are divisible by $3$ (sum of digits $1+3+6+5=15$). $$P = \frac{24}{455}$$ ### Exam Strategy & Shortcut When faced with large denominators like $1365$, don't calculate it immediately. Leave it as $15 \times 7 \times 13$ from the combination expansion. Then your fraction is $\frac{2 \times 6 \times 6}{15 \times 7 \times 13}$. The $15$ in the denominator and a $3$ from one of the $6$s will cancel. ### Common Pitfall Miscounting the number of marbles to draw. Ensure the sum of specific colors drawn ($1+2+1=4$) matches the total drawn in the denominator ($^NC_4$). ### Final Answer Therefore, the correct answer is **$\frac{24}{455}$**.