PEA 2024 PEA Q15 | ISI MSQE

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2024 PEA Q15

In the previous question, suppose in every period the monkey can go to any of the possible positions in the next period with equal probability. Then what is the probability that the monkey is at a distance of more than $1$ from $(0,0)$ in period $2$ ?

Reveal answer and solution

Answer

Solution

1
Each of the $9$ neighbours of $(0,0)$ is equally likely (probability $1/9$ ). Distances from $(0,0)$ :
2
\begin{itemize}[leftmargin=*]
3
- $(0,0)$ : distance $0$ .
4
- $(\pm 1,0),(0,\pm 1)$ : distance $1$ ( $4$ positions).
5
- $(\pm 1,\pm 1)$ : distance $\sqrt{2}>1$ ( $4$ positions).
6
\end{itemize}
7
Therefore $P(\text{distance}>1)=\dfrac{4}{9}$ .

Answer structure / marking notes

``Distance $>1$ '' is strict, so distance $=1$ does not count.

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Content note

Imported from public/resources/isi/msqe/solutions/pea/2024/ISI_MSQE_PEA_2024_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.