In a T-maze, a rate is given food if it turns left and an electric shock if it t
ID: 2983421 • Letter: I
Question
In a T-maze, a rate is given food if it turns left and an electric shock if it turns right. On the rst
trial there is a 50-50 chance that a rat will turn either way; then, if it receives food on a given trial
the probability that it will turn left on the next trial is 0.72, and if it receives a shock on a given trial
the probability that it will turn left on the next trial is 0.84.
1. What is the probability that the rat will turn left on the second trial?
2. Assuming the probability of a rat turning left on any given trials depends only on what it did
in the preceding trial, nd the probability that a rate will turn left on the third trial.
3. Removing the assumption in 2, calculate the expected number of trials ( in terms of expected value) that need to be made
until the rat turns right.
4. Removing the assumption from 2 again, calculate the expected number of trials ( in terms of expected value) needed until the rat turns right two times in a row.
Explanation / Answer
Hi, I am using Ri and Li to stand for turningright and left respectively on ith trial.
We are given:P(R1) = 0.5;
P(L1) = 0.5;P(L2 | L1) = 0.68;
from which we can infer
P(R2 | L1) = 1 - P(L2 |L1) = 1-0.68 = 0.32.
P(L2 | R1) = 0.84;
from which we can infer P(R2 | R1) = 1-P(L2 |R1) =1-0.84 = 0.16.
Therefore,
a) P(R2) = P(R2 | L1) * P(L1) + P(R2 | R1) * P(R1) {totalprobability theorem}
= 0.32 * 0.5 + 0.16 * 0.5
= 0.24.