Sammy has a counter that is red on one side and green on the other. The counter is shown below: A circular counter is shown. The top surface of the counter is shaded in a lighter shade of gray and Red is written across this section. The bottom section of the counter is shaded in darker shade of gray and Green is written across it. Sammy flips this counter 20 times. What is the probability that the 21st flip will result in the counter landing red side up? fraction 1 over 2 fraction 1 over 4 fraction 20 over 21 fraction 1 over 20

Question

Romeo Harmon

Mathematics

30 March, 18:57

Sammy has a counter that is red on one side and green on the other. The counter is shown below: A circular counter is shown. The top surface of the counter is shaded in a lighter shade of gray and Red is written across this section. The bottom section of the counter is shaded in darker shade of gray and Green is written across it. Sammy flips this counter 20 times. What is the probability that the 21st flip will result in the counter landing red side up? fraction 1 over 2 fraction 1 over 4 fraction 20 over 21 fraction 1 over 20

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Teagan Horne · Answer 1

The probability is 1/2 because there is a 50% chance it will land on either side. It does not matter how many times the coin has been flipped already.