Andy the Ant's Adventure

Let $X_i$ be the forward distance in meters that Andy travels during the $i$ th minute and note that

$E(X_i) = \frac{3}{4}\cdot (1) + \frac{1}{4} \cdot (-1) = \frac{1}{2}$

for all $i.$ Then, the forward distance Andy has traveled after one hour is simply $X_1+X_2+\ldots+X_{60}.$ By linearity of expectation,

$E\left[\sum_{i=1}^{60}X_i\right] = \sum_{i=1}^{60} E\left[X_i\right] = 60 \cdot \frac{1}{2} = 30.$

Because Andy moves left with probability 1/4 and right with probability 3/4 every minute, then you can expect Andy to move left 1 time and right 3 times every 4 minutes (just by multiplying by 4). This is a net change of 2 units to the right. Then, you can calculate that there are 15 groups of 4 minutes in an hour, so we multiply the net change of 2 units by 15 for a total of 30 units.

In one minute, it move 3/4 unit forward & 1/4 unit back. So only 1/2 unit. In 60 minutes it moves 60 x 1/2 = 30 units.

each time he back half from right to left so after 60 mint he will reach to 30 unit