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A frog starts at point A on the 3 × 3 grid of points shown below, where every pair of points that are 1 unit apart is connected by an edge. Every second, the frog randomly chooses an edge adjacent to its current point, and moves 1 unit along that edge. The frog may revisit an edge or point. What is the probability that after four seconds, the frog is at point B? Express your answer as a common fraction.

 

 Jan 5, 2019
 #1
avatar+101778 
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Look  at the grid below

 

7    8      9

 

4    5     6

 

1    2     3

 

First.....note that only  the corner points  and the middlle point can be reached in 4  moves from 1....so points 2, 4 , 6 and 8 cannot be reached in any set of 4 moves from 1

 

To reach 1 from 1 we have the following sets of moves

(1, 2, 1, 2, 1)  ( 1, 2, 5, 4, 1) (1, 4, 5, 2, 1) (1, 2, 3, 2, 1) (1, 4, 1, 4, 1) (1, 4, 7 , 4, 1)

To reach 3 from 1, we have these sets of moves

(1, 2, 3, 2, 3) ( 1, 2, 1, 2, 3 ) (1, 2, 3, 6, 3) (1, 2, 5, 6, 3) (1, 2, 3, 2, 1) ( 1, 4, 5, 6, 3) 

To reach 7 from 1 we have the same number of sets of moves as that of reaching 3 from 1

To reach the middle point 5 from 1 we have these sets of moves

(1, 2, 3, 6 , 5) (1, 2, 1, 2 , 5) (1, 2, 3, 2, 5) ( 1, 2, 5, 4 , 5) (1, 2, 5, 8, 5) (1, 2, 5, 2, 5) (1, 2, 5, 6, 5)

(1, 4, 7, 4, 5) ( 1, 4, 7, 8, 5) (1, 4, 5, 6, 5) (1, 4, 5, 8, 5) ( 1, 4, 1, 4 , 5)

To reach 9 from 1 we have

(1, 2, 3, 6, 9) (1, 2, 5, 6, 9) ( 1, 2, 5, 8, 9) (1, 4, 5, 6 , 9) (1, 4, 5, 8, 9) (1, 4, 7 , 8 , 9)

 

So.....we have a total of  4(6) + 12  =  36 possible moves

 

And 6 of these reach 9 from 1   -  B from A -.....so....the probability is  6 / 36   =  1 / 6

 

 

cool cool cool

 Jan 6, 2019

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