Five roommates

Five roommates will move into a house with four bedrooms: one double room and three single rooms. The five roommates propose that they draw names to determine the order in which they pick the bedrooms, assuming that the first three names drawn will choose the single rooms, so the two roommates who are not selected will need to share the double bedroom. Oddly, three of the guys get to enter their names thrice, while the other two only enter their names once (e.g., Alex, Bradley, and Carter each get to submit their names three times, while Derek and Edward only get to submit their names once); this produces a total of 11 entries in the drawing. Once someone has been drawn, any duplicates of his name are removed from the pool. What is the probability that Carter is one of the first three names drawn (excluding duplicates) and can choose a single room?

Correct answer:

p =  0.8571

Step-by-step explanation:

$$\begin{gathered} n=5 \\ A=3 \\ B=3 \\ C=3 \\ D=1 \\ E=1 \\ t=A+B+C+D+E=3+3+3+1+1=11 \\ {c}_1=\frac{C}{t}=\frac{3}{11}\doteq 0.2727 \\ {c}_2=(1-c_1)\cdot \frac{C}{B+C+D+E}=(1-\frac{3}{11})\cdot \frac{3}{3+3+1+1}=\frac{3}{11}\doteq 0.2727 \\ {c}_3=(1-c_2)\cdot \frac{C}{B+C+D}=(1-\frac{3}{11})\cdot \frac{3}{3+3+1}=\frac{24}{77}\doteq 0.3117 \\ p=c_1+c_2+c_3=\frac{3}{11}+\frac{3}{11}+\frac{24}{77}=\frac{21}{77}+\frac{21}{77}+\frac{24}{77}=\frac{21+21+24}{77}=\frac{66}{77}=\frac{6}{7}=0.8571 \end{gathered}$$

Try another example