Core Connections Geometry, 2013
CC
Core Connections Geometry, 2013 View details
2. Section 3.2
Continue to next subchapter

Exercise 79 Page 182

Practice makes perfect
a We want to make a tree diagram of all the possible outcomes of spinning two spinners involved in a game show. The first spinner has three possible outcomes, which cause an initial three branches in our diagram. The second spinner has two outcomes, which means the first three branches all split into two branches.
b We want to decide whether each of the outcomes in the sample space is equally likely. First, let's think about what the sample space is in this example. To do so, let's look at the tree diagram from Part A.

The sample space is the set of all possible outcomes. If we have a tree diagram, all outcomes are represented as an end of a branch. Here we have 6 branches in total, so there are 6 possible outcomes. cc Sample Space:& {$ 200, $ 100, $ 600, & $ 300, $ 3000, $ 1500}

Assuming that the sectors on the first spinner are equally large and that the sectors on the second spinner are equally large, the probability of each of the 6 branch endings of our diagram is the same, 1 6. Since each branch ending corresponds to a different outcome, each outcome is equally likely.
c There is only one outcome that results in winning $200. First, we have to win $100 on the first wheel, and then we have to double that amount on the second wheel.
As already explained in Part B, the probability of each outcome is 16. Therefore the probability of taking home $200 is 16. To find the probability of taking home more than $500, let's highlight the outcomes where the prize is more than $500.

There are three outcomes that result in winning more than $500. P($300, double)= 16 P($1500, double)= 16 P($1500, keep)= 16 To calculate the probability that we get one of these options, we have to add their probabilities. P(more than$500): 1/6+1/6+1/6=3/6 The probability of winning more than $500 is 36, which can be reduced to 12.

d The probability of doubling the winnings depends only on the outcome of the second spin. This is because no matter what the result of the first spin is, the second spinner remains the same. From the second spinner, we see that each field occupies half the spinner. Therefore, the probability of doubling our winnings is 12.
e Let's rework the tree diagram with the new amounts and highlight the paths that result in winning $200.

Now there are two outcomes that result in winning $200. P($100, double)=1/6 P($200, keep)=1/6 Like in Part C, to calculate the probability that we get either of these outcomes we have to add their probabilities. P($200): 1/6+1/6=2/6 The probability of winning $200 is 26, which can be reduced to 13.