CMI2018-A-7

Question

Let $C_{n}$ be the number of strings $w$ consisting of $n$ $X's$ and $n$ $Y's$ such that no initial segment of $w$ has more $Y's$ than $X's.$ Now consider the following problem. A person stands on the edge of a swimming pool holding a bag of $n$ red and $n$ blue balls. He draws a ball out one at a time and discards it. If he draws a blue ball, he takes one step back, if he draws a red ball, he moves one step forward. What is the probability that the person remains dry?

• A. $\frac{C_{n}}{2^{2n}}$
• B. $\frac{C_{n}}{\binom{2n}{n}}$
• C. $\frac{n\cdot C_{n}}{(2n)!}$
• D. $\frac{n\cdot C_{n}}{\binom{2n}{n}}$

Answer 1

$\mathbf{\underline{Answer:}\Rightarrow} \mathbf{B}$

$\mathbf{\underline{Explanation:}\Rightarrow} \mathbf{}$

Total ways in which $\color {green} {\Large {2\mathrm n}}$ balls can be chosen where $\color {red} {\mathrm {\Large n}}$ are of $ \color {red} {\text{Red Colour}}$ and other $\color {blue}{\Large {\mathrm n}}$ are of $\color {blue} {\text{Blue Colour}}$ = $\dfrac{(2\mathrm n)!}{\mathrm n!.\mathrm n!}$ = $\dfrac{2}{\mathrm n!}$ = $\Large \binom{2\mathrm n}{\mathrm {C_n}}$

For the person to remain dry he should have chosen more Red balls than the blue balls which is given by $\Large \color {darkgreen} {\mathrm {C_n}}$.

$\therefore$ The answer $=\color {green} {\Large {\dfrac{\mathrm {C_n}}{\binom{2\mathrm n}{\mathrm {C_n}}}}}$

$\therefore $ B is the correct option.

Answer 2