Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
1. Section 3.1
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Exercise 32 Page 159

Practice makes perfect
a 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 will all split into two branches.
b Looking at the diagram, we see that the sectors on the first spinner are equally large and that the sectors on the second spinner are equally large. Therefore, the probability of each of the six outcomes in the tree diagram will be the same, 16.
c Let's look at our diagram. We can see that 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, so the probability of taking home $200 is 16. Let's highlight the outcomes where we win more than $500.

There are three outcomes that result in winnings of more than $500. P($300, double)= 16 P($1500, double)= 16 P($1500, keep)= 16 To calculate the probability that we spin any 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 on the second spinner only. From the second spinner, we see that each field occupies half of 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 winnings of $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 you spin either option 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.