Question

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 aleast $2$ of them goes in hotel $x$, atleast $1$ in hotel $y$ and atleast $1$ in hotel $z$? (each hotel has capacity for more than $20$ guests).

• A. $\frac{{}^{18}{C}_2}{{}^{22}{C}_2}$
• B. $\frac{{}^{20}{C}_2{.}^{18}{C}_1{.}^{17}{C}_1.3^{16}}{3^{20}}$
• C. $\frac{{}^{20}{C}_9}{3^9}$
• D. $\frac{3^{20}-{13.2}^{20}+43}{3^{20}}$

Answer 1

The correct option is A $\frac{{}^{18}{C}_2}{{}^{22}{C}_2}$

There are $3$ hotels $(x,y,z)$ $\xi$ $20$ people.

Let $x_1,x_2,x_3$ represent the no of people in $x,y,z$ respectively.

$x_1+x_2+x_3=20$ $,$ $x_i\ge 0$

No. of solutions $={}^{n+r-1}{C}_{r-1}$

Here $n=20,r=3$

$\therefore$ Total possible ways $={}^{20+3-1}{C}_{3-1}={}^{22}{C}_2$

Now, they have give that $x_1\ge 2,x_2\ge 1,x_3\ge 1$

Let $x_1-1=y_1,x_1-2\ge 0\to y_1\ge 1$

$x_2=y_2$ $,$ $x_3=y_3$ $,$ $y_2,y_3\ge 1$

$\therefore$ The equation becomes $(y_1+1)+y_2+y_3=20$

$y_1+y_2+y_3=19$ $,$ $y_i\ge 1$

No. of solutions $={}^{n-1}{C}_{r-1}.$

Here $n=19,r=3$

Required $={}^{18}{C}_2$

Probability $=\frac{req,}{total}=\frac{{}^{18}{C}_2}{{}^{22}{C}_2}.$

Hence, the answer is $\frac{{}^{18}{C}_2}{{}^{22}{C}_2}.$