# Video: AF5P2-Q11-648126135175

A fair spinner has 20 equal sections. Label each section π΄, π΅, πΆ, or π· so that when the arrow is spun, the probability it lands on π΄ is 1/5, the probability it lands on πΆ is the probability it lands on π΄ or π΅, and the probability it lands on π΅ is half the probability it lands on π΄.

02:56

### Video Transcript

A fair spinner has 20 equal sections. Label each section π΄, π΅, πΆ, or π· so that when the arrow is spun, the probability it lands on π΄ is one-fifth, the probability it lands on πΆ is the probability it lands on π΄ or π΅, and the probability it lands on π΅ is half the probability it lands on π΄.

The first piece of information weβre given says the probability that the spinner lands on π΄ is one-fifth. If the probability of π΄ 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 π΄. What do we know about the probability of π΅? The probability of π΅ is half of the probability of π΄. One-half times the probability of π΄ 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 π΅ is two twentieths. This makes sense. If we labeled four of the spaces with π΄ and the probability it lands on π΅ is half, then we would label two spaces with π΅. After that, we have the probability of πΆ. The probability of πΆ is the probability that it lands on π΄ or π΅. This would be four twentieths plus two twentieths. The probability that it lands on πΆ is six twentieths. And that means we need to label six of the sections with πΆ. Again, this makes sense. If πΆ is the probability of π΄ or π΅, together π΄ or π΅ makes up six of the spaces on our spinner.

Now we filled π΄, π΅, and πΆ. We know that the spinner also has the letter π·. Weβre not given any additional information. And so weβll fill the rest of the spinner with the letter π·. There are eight remaining spaces. And so we can say the probability of landing on π· is eight out of 20. The correct spinner will have four π΄βs, two π΅βs, six πΆβs, and eight π·βs. These letters could be placed anywhere on the spinner as long as you have the correct number of each letter.