Question

Twenty persons arrive in a town having 3 hotels A, B and C. If each person randomly chooses one of these hotels, then what is the probability that atleast two of them go in hotel A, atleast one in hotel B and atleast one in hotel C. (each hotel has capacity for more than 20 guests)

• A. 1-($\frac{12.2^{20}-42}{3^{20}}$)
• B. 1-($\frac{2^{20}-1}{3^{19}}$)
• C. 1-($\frac{10.2^{20}-40}{3^{20}}$)
• D. 1-($\frac{13.2^{20}-43}{3^{20}}$)

Answer 1

Answer: (D)

1-$\left(\frac{13.2^{20}-43}{3^{20}}\right)$

Answer 2

The total number of ways to distribute 20 persons into 3 hotels is $3^{20}$. Let $E$ be the event that at least two persons go to hotel A ($n_A \ge 2$), at least one to hotel B ($n_B \ge 1$), and at least one to hotel C ($n_C \ge 1$). It is easier to calculate the probability of the complement event $E^c$. $E^c$ is the event that ($n_A < 2$) OR ($n_B = 0$) OR ($n_C = 0$). Let $C_A$ be $n_A < 2$, $C_B$ be $n_B = 0$, and $C_C$ be $n_C = 0$. Using the Principle of Inclusion-Exclusion: $|E^c| = |C_A| + |C_B| + |C_C| - (|C_A \cap C_B| + |C_A \cap C_C| + |C_B \cap C_C|) + |C_A \cap C_B \cap C_C|$.

• $|C_B| = 2^{20}$ (persons choose A or C).
• $|C_C| = 2^{20}$ (persons choose A or B).
• $|C_A|$: ($n_A=0$ or $n_A=1$).
  • $n_A=0$: $2^{20}$ ways (choose B or C).
  • $n_A=1$: $\binom{20}{1} \times 2^{19}$ ways (choose 1 for A, rest for B or C). $|C_A| = 2^{20} + 20 \times 2^{19} = 2^{20} + 10 \times 2^{20} = 11 \times 2^{20}$.
• $|C_B \cap C_C|$: $n_B=0$ and $n_C=0 \implies n_A=20$. Only 1 way.
• $|C_A \cap C_B|$: ($n_A < 2$) and ($n_B=0$). This means ($n_A=0, n_B=0$) or ($n_A=1, n_B=0$).
  • $n_A=0, n_B=0 \implies n_C=20$: 1 way.
  • $n_A=1, n_B=0 \implies n_C=19$: $\binom{20}{1} \times 1 = 20$ ways. $|C_A \cap C_B| = 1 + 20 = 21$.
• $|C_A \cap C_C|$: Similarly, $|C_A \cap C_C| = 21$.
• $|C_A \cap C_B \cap C_C|$: ($n_A < 2$) and ($n_B=0$) and ($n_C=0$). This implies $n_A=20$, which contradicts $n_A < 2$. So, 0 ways. $|E^c| = (11 \times 2^{20}) + 2^{20} + 2^{20} - (21 + 21 + 1) + 0 = 13 \times 2^{20} - 43$. $P(E^c) = \frac{13 \times 2^{20} - 43}{3^{20}}$. $P(E) = 1 - P(E^c) = 1 - \frac{13 \times 2^{20} - 43}{3^{20}}$.