# Video: AF5P2-Q11-648126135175

A fair spinner has 20 equal sections. Label each section 𝐴, 𝐵, 𝐶, or 𝐷 so that when the arrow is spun, the probability it lands on 𝐴 is 1/5, the probability it lands on 𝐶 is the probability it lands on 𝐴 or 𝐵, and the probability it lands on 𝐵 is half the probability it lands on 𝐴.

02:56

### Video Transcript

A fair spinner has 20 equal sections. Label each section 𝐴, 𝐵, 𝐶, or 𝐷 so that when the arrow is spun, the probability it lands on 𝐴 is one-fifth, the probability it lands on 𝐶 is the probability it lands on 𝐴 or 𝐵, and the probability it lands on 𝐵 is half the probability it lands on 𝐴.

The first piece of information we’re given says the probability that the spinner lands on 𝐴 is one-fifth. If the probability of 𝐴 is one-fifth and there are 20 spaces, we know that five times four equals 20. And if we multiply the denominator by a number, we need to multiply the numerator by the same number. one-fifth is equal to four twentieths. And that means four of the spaces need to be labeled with an 𝐴. What do we know about the probability of 𝐵? The probability of 𝐵 is half of the probability of 𝐴. One-half times the probability of 𝐴 equals one-half times four twentieths.

If we simplify this multiplication, we can divide the numerator and the denominator by two. And we see that the probability of 𝐵 is two twentieths. This makes sense. If we labeled four of the spaces with 𝐴 and the probability it lands on 𝐵 is half, then we would label two spaces with 𝐵. After that, we have the probability of 𝐶. The probability of 𝐶 is the probability that it lands on 𝐴 or 𝐵. This would be four twentieths plus two twentieths. The probability that it lands on 𝐶 is six twentieths. And that means we need to label six of the sections with 𝐶. Again, this makes sense. If 𝐶 is the probability of 𝐴 or 𝐵, together 𝐴 or 𝐵 makes up six of the spaces on our spinner.

Now we filled 𝐴, 𝐵, and 𝐶. We know that the spinner also has the letter 𝐷. We’re not given any additional information. And so we’ll fill the rest of the spinner with the letter 𝐷. There are eight remaining spaces. And so we can say the probability of landing on 𝐷 is eight out of 20. The correct spinner will have four 𝐴’s, two 𝐵’s, six 𝐶’s, and eight 𝐷’s. These letters could be placed anywhere on the spinner as long as you have the correct number of each letter.