Question

OMG PLS HELP

2. Your pet rat Randy is the next contestant on a new game show What a Rat! Where

rats compete in mazes to find food as

fast as possible. Randy is equally

likely to take any path when coming

to an intersection, but you get to

decide which area to place the food.

Given the maze above, in which area

should you place the food to help

Randy win the competition? Show all

of your calculations! Be clear,

complete and justify your answer.

Answer 1

Answer:

Box C

Step-by-step explanation:

Given

See attachment 1 for the maze

Required

The area to place the food in order to help the rat win

To answer this question, we make use of the concept of probability (see attachment 2)

From the starting point to the first joint:

Each of the three ways have the same probability of 1/3

1. If the car passes the way at the top, there are two additional joints.

1.1 Each of the two joints (top and bottom) have the same probability of 1/6 (i.e. divide 1/3 by 2)

1.2 The bottom joint also splits to two (each of these ways have a probability of 1/12; i.e. divide 1.6 by 2)

2. If the car passes the way at the middle, there are two additional joints.

Each of the two joints (top and bottom) have the same probability of 1/6 (i.e. divide 1/3 by 2)

3. If the car passes the way at the bottom, there are three additional joints.

Each of the three joints have the same probability of 1/9 (i.e. divide 1/3 by 9)

Next, is to sum up the probability of each box.

For Box A

$P(A) = \frac{1}{6} + \frac{1}{12} + \frac{1}{9}$

Take LCM

$P(A) = \frac{6+3+4}{36}$

$P(A) = \frac{13}{36}$

$P(A) = 0.361$

For Box B

$P(B)= \frac{1}{12} + \frac{1}{6}$

Take LCM

$P(B)= \frac{1+2}{12}$

$P(B)= \frac{3}{12}$

$P(B)= 0.250$

For Box C

$P(C) = \frac{1}{6} + \frac{1}{9} + \frac{1}{9}$

Take LCM

$P(C) = \frac{3+2+2}{18}$

$P(C) = \frac{7}{18}$

$P(C) = 0.389$

From the calculations above, box C has the highest probability of 0.389 compared to 0.250 (of box B) and 0.361 (of box A)

Hence, the food should be placed in Box C