Terence has a checkerboard with 10 rows and 10 columns, a single checker at the bottom-left corner, and a spinner with the numbers 1,2, and 3. There is an equal chance of each number being spun.
He first spins the spinner and moves the checker from the bottom-left corner up the number of spaces spun. He then spins the spinner again, this time moving the checker right the number
of spaces spun.
He repeats this process twice more, thus spinning the spinner for a total of six times from the outset.
Determine the probability that the checker will land on the same color as the square on which it started.
First note that the order of the turns doesn't matter. Essentially you will spin three times to see how far up the board to move, and three times to see how far to the right.
If the checker moves up the board an even number of squares, it would need to move an even number of squares to the right to land on the same color as the square on which it started.
Similarly if it moves up an odd number of squares, it needs to move an odds number of squares to the right.
If one sum is even and the other is odd, on the other hand, it will land on the opposite color.
So basically if the sum of all six spins is even, the checker will land on the same color as the square on which it started. If it's odd, it will not.
The probability that the sum will be even is 365 / 729.
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Posted by tomarken
on 2022-02-07 08:26:33 |