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(D) Let $n_x, n_y, n_z$ be the number of persons in hotels $x, y, z$ respectively. We have $n_x + n_y + n_z = 20$. We want to find the number of ways

Vedclass · Accepted Answer

$\frac{{3^{20} - 13 \cdot 2^{20} + 43}}{{3^{20}}}$

Vedclass · Answer

(D) Let $n_x, n_y, n_z$ be the number of persons in hotels $x, y, z$ respectively. We have $n_x + n_y + n_z = 20$. We want to find the number of ways such that $n_x \geq 2, n_y \geq 1, n_z \geq 1$.Total ways = $3^{20}$.Let $S$ be the total set of outcomes. Let $A$ be the condition $n_x $n(B) = 2^{20}$ (all in $x$ or $z$),$n(C) = 2^{20}$ (all in $x$ or $y$).$n(A) = n(n_x=0) + n(n_x=1) = 2^{20} + 20 \cdot 2^{19}$.Using the Principle of Inclusion-Exclusion for the complement set (where the condition is $NOT$ met),the number of favorable cases is $3^{20} - [n(A) + n(B) + n(C) - (n(A \cap B) + n(B \cap C) + n(C \cap A)) + n(A \cap B \cap C)]$.After calculating the intersections and applying the formula,the number of favorable cases is $3^{20} - 13 \cdot 2^{20} + 43$.Thus,the probability is $\frac{3^{20} - 13 \cdot 2^{20} + 43}{3^{20}}$.

Twenty persons arrive in a town having $3$ hotels $x, y$ and $z$. If each person randomly chooses one of these hotels,then what is the probability that at least $2$ of them go to hotel $x$,at least $1$ to hotel $y$,and at least $1$ to hotel $z$? (Each hotel has a capacity for more than $20$ guests.)

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