ProbabilityHardBloom L3

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)

Options

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

Solution

Sol. [Key Concept: $20$ persons into 3 hotels: complementary count (each hotel non-empty).]

A = number of persons going to hotel $A = 1$

$B =$ number of persons going to hotel $A = 0$

$C =$ number of persons going to hotel $B = 0$

$D =$ number of persons going to hotel $C = 0$

$$P = 1 - \frac{n(\text{ }A \cup \text{ }B \cup C \cup D)}{n(\text{ }s)}$$

$$\begin{matrix} n(\text{ }A \cup \text{ }B \cup C \cup D) & \ = \sum_{}^{}\ n(\text{ }A) - \sum_{}^{}\ n(\text{ }A \cap \text{ }B) + \sum_{}^{}\ n(\text{ }A \cap \text{ }B \cap C) - n(\text{ }A \cap \text{ }B \cap C \cap D) \\ & \ = \left( \ ^{20}C_{1}2^{19} + {3.2}^{20} \right) - \left( \ ^{20}C_{1}1^{19} \times 2 + 1 \times 2 + 1 \right) \\ \ & \\ \therefore & \ P = 1 - \left( \frac{{13.2}^{20} - 43}{3^{20}} \right) \end{matrix}$$

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