Question

An ant is situated at the vertex A of the triangle ABC. Every movement of the ant consists of moving to one of other two adjacent vertices from the vertex where it is situated. The probability of going to any of the other two adjacent vertices of the triangle is equal. The probability that at the end of the fourth movement the ant will be back to the vertex A, is :

• A. $\frac{4}{16}$
• B. $\frac{6}{16}$
• C. $\frac{7}{16}$
• D. $\frac{8}{16}$

Answer 1

Answer: (A)

$\frac{4}{16}$

$Given$: Ant is initially at vertex $A$.

Probabilty of ant going to each of two way from a vertex $=$ $\frac{1}{2}$ [As, both are equally likely]

Now, we observe that there are four possible ways in which ant will return to vertex A in fourth movement as follows:

1. $AB\to BC\to CB\to BA$

2. $AC\to CB\to BC\to CA$

3. $AB\to BA\to AC\to CA$

4. $AC\to CA\to AB\to BA$

Probability of event $1$. $=$ $\left(\frac{1}{2}\right)\times \left(\frac{1}{2}\right)\times \left(\frac{1}{2}\right)\times \left(\frac{1}{2}\right)$

$=$ $\frac{1}{16}$

Similarly, Probability of all four event are same.

$\therefore$ Total probability of favourable 4 events= $\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}\Rightarrow \frac{4}{16}$

Hence, Option (A) is correct.