Steven has a counter that is yellow on one side and blue 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 Yellow is written across this section. The bottom section of the counter is shaded in darker shade of gray and Blue is written across it.

Steven flips this counter 22 times. What is the probability that the 23rd flip will result in the counter landing on yellow side up?

fraction 22 over 23
fraction 1 over 2
fraction 1 over 4
fraction 1 over 22

Since it's a coin flip, each coin flip is independent of any other. Each time Steven flips the counter it has the same chance of landing on either side. It's true the on the first throw, and it's true on the 1000th throw.

Even if, extremely improbable, the first 22 flips landed on the blue side, that predicts nothing regarding the outcome of the 23rd flip.

That's all assuming the counter/chip/coin is properly and evenly balanced.... since we have no indication otherwise in the problem definition, we have to assume it's the case.

williehamm914 williehamm914 · Answer 2 · 2022-04-18T19:43:20+02:00

williehamm914 williehamm914

18-04-2022

Answer:

I know im pretty late (2022!) But the correct answer is B. Fraction 1 over 2([tex]\frac{1}{2}[/tex] This is what it would look like as shown In the other answer.)

Step-by-step explanation: If you would look in the picture you wold see Halve blue and halve yellow as well. Proof is Flvs 7th grade math.

Thank you "Halve" A wonderful day! Haha and have a happy one day late Easter!

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