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There are two rocks $A$ and $B$, located close to each other, in a lily pond. There is a frog that jumps randomly between the two rocks at time $t = 0,1, 2, \ldots$. The location of the frog is determined as follows. Initially, at time $t = 0$, the frog is at $A$. From then on, the frog's location is determined as follows. If the frog is at $A$ at time $t$, then at time $t + 1$, with probability $2/3$ it jumps to $B$ and with probability $1/3$, it jumps on the spot and stays at $A$. If the frog is at $B$ at time $t$, then at time $t + 1$, with probability $1/2$ it jumps to $A$ and with probability $1/2$ it jumps on the spot and stays at $B$. What is the probability that the frog is at $B$ at time $3$ (just after its third jump)?

- $\frac{1}{2}$
- $\frac{31}{54}$
- $\frac{14}{27}$
- $\frac{61}{108}$
- $\frac{2}{3}$

18 votes

Best answer

In such cases where the events are mutually exclusive and collectively exhaustive, it is preferable to use tree diagram to find the final outcome. The tree diagram for the given problem is as shown:

The frog is at $B$ at $t = 3$

$\implies P(A .A.B ) + P(A.B.B) + P(B.A.B) + P(B.B.B)$

$=\frac{1}{3}\times \frac{1}{3}\times \frac{2}{3} + \frac{1}{3}\times \frac{2}{3}\times \frac{1}{2} + \frac{2}{3}\times \frac{1}{2}\times \frac{2}{3} + \frac{2}{3}\times \frac{1}{2}\times \frac{1}{2}$

$=\frac{2}{27} + \frac{1}{9} + \frac{2}{9} + \frac{1}{6} = \frac{31}{54}$

Hence we have $P($frog is at $B$ at $t = 3) = 31/54$

Correct Answer: $B$