csat-suneung· South-Korea· csat__math-science4 marks
From $3n$ cards labeled with the numbers $1,2,3 , \cdots , 3 n$ ($n$ is a natural number), two cards are randomly drawn. Let the two numbers on the cards be $a , b$ ($a < b$) respectively. Let $\mathrm { P } _ { n }$ be the probability that $3 a < b$. The following is the process of finding $\lim _ { n \rightarrow \infty } \mathrm { P } _ { n }$. The number of ways to draw 2 cards from $3n$ cards is ${ } _ { 3 n } \mathrm { C } _ { 2 }$. For $3 a < b$, we have $b \leqq 3 n$, so $1 \leqq a < n$. Therefore, if $a = k$, the number of cases of $b$ satisfying $3 a < b$ is (A), so $$\mathrm { P } _ { n } = \frac { \text{(B)} } { { } _ { 3 n } \mathrm { C } _ { 2 } } \text{.}$$ Therefore, $\lim _ { n \rightarrow \infty } \mathrm { P } _ { n } =$ (C). What are the correct values for (A), (B), and (C) in the above process? [4 points] (1) (A) $3(n-k)$, (B) $\frac{3}{2}n(n-1)$, (C) $\frac{1}{3}$ (2) (A) $3(n-k)$, (B) $\frac{3}{2}n(n-1)$, (C) $\frac{2}{3}$ (3) (A) $3(n-k)$, (B) $3n(n-1)$, (C) $\frac{2}{3}$ (4) (A) $3(n-k+1)$, (B) $3n(n-1)$, (C) $\frac{1}{3}$ (5) (A) $3(n-k+1)$, (B) $3n(n-1)$, (C) $\frac{2}{3}$
From $3n$ cards labeled with the numbers $1,2,3 , \cdots , 3 n$ ($n$ is a natural number), two cards are randomly drawn. Let the two numbers on the cards be $a , b$ ($a < b$) respectively. Let $\mathrm { P } _ { n }$ be the probability that $3 a < b$. The following is the process of finding $\lim _ { n \rightarrow \infty } \mathrm { P } _ { n }$.
The number of ways to draw 2 cards from $3n$ cards is ${ } _ { 3 n } \mathrm { C } _ { 2 }$.
For $3 a < b$, we have $b \leqq 3 n$, so $1 \leqq a < n$.
Therefore, if $a = k$, the number of cases of $b$ satisfying $3 a < b$ is (A), so
$$\mathrm { P } _ { n } = \frac { \text{(B)} } { { } _ { 3 n } \mathrm { C } _ { 2 } } \text{.}$$
Therefore, $\lim _ { n \rightarrow \infty } \mathrm { P } _ { n } =$ (C).
What are the correct values for (A), (B), and (C) in the above process? [4 points]\\
(1) (A) $3(n-k)$, (B) $\frac{3}{2}n(n-1)$, (C) $\frac{1}{3}$\\
(2) (A) $3(n-k)$, (B) $\frac{3}{2}n(n-1)$, (C) $\frac{2}{3}$\\
(3) (A) $3(n-k)$, (B) $3n(n-1)$, (C) $\frac{2}{3}$\\
(4) (A) $3(n-k+1)$, (B) $3n(n-1)$, (C) $\frac{1}{3}$\\
(5) (A) $3(n-k+1)$, (B) $3n(n-1)$, (C) $\frac{2}{3}$