Chapter Closure
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Exercise 190 Page 665
a Add the probabilities of moving 4 and 6 spaces from the two spinners.
b There is only one region of one spinner where Giancarlo stays put.
c Consider the complement to your answer in Part B.
d Calculate P(S2 | 2 spaces).
e Consider the formula P(A | B)= P(A and B)P(B).
a About 63.3 %
b About 16.7 %
c About 83.3 %
d About 58.8 %
e See solution.
a From the exercise we know that S1 decides which of S2 and S3 we should spin. There is a 23 probability of spinning S2 and 13 of spinning S3. With this information we can begin drawing the two-way table.
S2
Examining S2, we notice that each region occupies a quarter of the circle. Therefore, we have an equal probability of spinning 4 and 6 spaces.P(4 spaces)=1/4 [0.5em] P(6 spaces)=1/4
S3
Examining S3, we can determine the probability of spinning 4 and 6 spaces. P(4 spaces)=2/5 [0.5em] P(6 spaces)=1/4
Two-Way Table
When we know the probability of the different regions we can use the Multiplication Rule of Probability to calculate the probability of first getting to spin each of the spinners and then the relevant regions. rl P(S2 and 4 spaces):& 2/3* 1/4=2/12 [1.2em] P(S2 and 6 spaces):& 2/3* 1/4=2/12 [1em] [-0.7em] P(S3 and 4 spaces):& 1/3* 2/5=2/15 [1.2em] P(S3 and 6 spaces):& 1/3* 1/4=1/12 Let's add the probabilities to the two-way table.
The probability of moving 4 or 6 spaces is the sum of the probabilities in the two-way table. P(4 or 6): 2/12+2/12+2/15+1/12≈ 63.3 %
b If he stays put, he has to spin a 0 on S2. As already explained in Part A, to calculate the probability of this outcome we have to multiply the probabilities of spinning S2 and then 0.
P(S2 and 0 spaces):& 2/3* 1/4=2/12 If we reduce the fraction we see that there is a 16 probability of staying put. This is the same thing as 16.7 %.
c Getting to move is the complement of staying put.
P(move)=1-P(stay put)
Substitute
P(stay put)= 1/6
P(move)=1- 1/6
▼
Simplify right-hand side
OneToFrac
Rewrite 1 as 6/6
P(move)=6/6-1/6
SubFrac
Subtract fractions
P(move)=5/6
UseCalc
Use a calculator
P(move)=0.83333...
WritePercent
Convert to percent
P(move)=83.3 %
d In this case it is given that Antonio moved 2 spaces. Therefore, we want to determine the probability of moving 2 spaces by spinning S2 given that we move 2 spaces. In other words, we want to divide the content of the blue rectangle with the content of the red rectangle below.
We know that the probability of spinning each region on S2 is 212. Let's add this information to the two-way table.
P(2 | S3): 1-2/5-1/4=7/20 Now we can calculate the probability of first spinning S3 and then a 2. P(S3 and 2): 1/3* 7/20 =7/60 Let's add this information to the diagram.
Now we have all the information we need to calculate the probability of spinning S2 on the first spinner given that he spun a 2 on the second spinner. P(S2 | 2 spaces): 212/2 12+ 760≈ 58.8 %
e Conditional probability means that we are looking for the probability of an event given that another event has happened.
P(A | B)=P(A and B)/P(B) In this case, P(A and B) refers to the probability of spinning S2 and getting to move 2 spaces while P(B) is the probability of getting to move 2 spaces in general.
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