Exercise 110 Page 262

Practice makes perfect

a Each spinner has three sectors that occupy a certain ratio of the circle. These ratios reflect their respective probability. In the first spinner, two sectors have angles of 90^(∘), which means the third sector has an angle of 180^(∘). In the second spinner, all sectors occupy a third of the circle. In other words, all of them have angles of 360^(∘)3=120^(∘).

Since we know the probability of obtaining each sector in our respective spinners, we can draw a tree diagram and highlight the outcome that lets Harvey be able to move his marker.

The probability of getting purple on both spinners is the product of the probabilities along this path in the tree diagram. P(purple on both) = 1/4*1/3=1/12

b The intersection of two events describes a situation where two events both happen. The union of two events describes a situation where either event can happen, but not necessarily both. In our case, we are looking for the probability of getting purple on both spinners. Therefore, the event that Harvey wins is an intersection.

c The best color is the choice that maximizes Harvey's probability of winning. The sectors of the second spinner are all equal.

Therefore, the choice of color does not matter here. However, in the first spinner yellow occupies half the circle, while purple only occupies a quarter.

Therefore, Harvey should choose yellow.

d Let's highlight all the paths that result in no one getting to move their marker.

We have six paths through the tree diagram where nobody can move their marker. We calculate the probability of each outcome by multiplying the probabilities along each path of the tree diagram. P(Green, Purple)& = 1/4*1/3=1/12 [0.8em] P(Green, Yellow)& = 1/4*1/3=1/12 [0.8em] P(Purple, Green)& = 1/4*1/3=1/12 [0.8em] P(Purple, Yellow)& = 1/4*1/3=1/12 [0.8em] P(Yellow, Green)& = 1/2*1/3=1/6 [0.8em] P(Yellow, Purple)& = 1/2*1/3=1/6 [0.8em] To calculate the probability that you get any of these events, we add their probabilities.

P(no one gets to move)=1/12*4 +1/6* 2

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Simplify right-hand side

MoveRightFacToNumOne

1/b* a = a/b

P(no one gets to move)=4/12+2/6

ExpandFrac

a/b=a * 2/b * 2

P(no one gets to move)=4/12+4/12

AddFrac

Add fractions

P(no one gets to move)=8/12

ReduceFrac

a/b=.a /4./.b /4.

P(no one gets to move)=2/3

e In Part D we found the probability of spinning different colors by calculating the probability of these combinations and adding them. However, we could also calculate the probability of getting the same color and then find the complement of this.

1-P(Same color) Let's highlight the paths in the tree diagram where you get to move your marker.

We have three paths through the tree diagram where a marker can be moved. We calculate the probability of each outcome by multiplying the probabilities along each path of the tree diagram. P(Green, Green)& = 1/4*1/3=1/12 [0.8em] P(Purple, Purple)& = 1/4*1/3=1/12 [0.8em] P(Yellow, Yellow)& = 1/2*1/3=1/6 To calculate the probability that any of these events happen, we add their probabilities.

P(you can move)=1/12+1/12+1/6

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