In a laboratory experiment an attempt is made to teach an animal to turn right in a maze. To aid in the teaching the animal is rewarded if it turns right on a given trial and punished if it turns left. On the first trial the animal is just as likely to turn right as left. If on a particular trial the animal was rewarded, his probability of turning right on the next trial is $p_1>\frac{1}{2}$, and if on a given trial the animal was punished his probability of turning right on the next trial is $p_2>p1$.
We need to find the probability that the animal will turn right on the third trial.
That's possible if the animal takes either of the following situations(added together):
$RRR+RLR+LRR+LLR$ (For example the first says, animal goes Right, Right and the again Right).
I found the answer to be: $(0.5)(p_1)(p_1)+(0.5)(1-p_1)(p_2)+(0.5)(p_2)(p_1)+(0.5)(1-p_2)(p_2)$
Sorry for not simplifying more but is that correct ?
Yes, your answer is perfectly correct! Simplifying it further gives us: $$P_{\text{req}} = 0.5p_1^2+0.5p_2-0.5p_1p_2+0.5p_1p_2+0.5p_2-0.5p_2^2$$ $$= 0.5\left[p_1^2-p_2^2\right]+p_2$$