Activity 4 Results

1. Record the number of each color of chips for each side of the container.

Red White Blue
Side A
Side B

Situation 3

Amos, Bertie, and Chloe are servers for Damon's Dinner Dive. Amos serves 40% of the customers and receives complaints from 3% of his customers. Bertie serves 35% of the customers and receives complaints from 4% of her customers. Chloe serves 25% of the customers and receives complaints from 6% of her customers.

14. Label the tree diagram appropriately.

15. Complete the joint probability distribution.

Yes No Total
Amos 0.012 0.388 0.400
Bertie 0.014 0.336 0.350
Chloe 0.015 0.235 0.250
Total 0.041 0.959 1.000

16. What percent of the customers complain about a server? 0.041 = 4.1%

17. A customer complains about a server. What is the probability the server was Chloe? 0.015 / 0.041 = 0.3659

Situation 4

A bag of candy has the following flavor distribution. 20% grape, 40% orange, and 30% cherry. The rest are lemon flavored. Assume that the bag has enough candy in it that the distribution is not affected by sampling without replacement.

18. Create a probability distribution for the flavor of candy.

Flavor Grape Orange Cherry Lemon Total
Probability 0.2 0.4 0.3 0.1 1.0

19. A single candy is randomly selected from the bag.

19a. What is the probability that it is lemon? 0.1

19b. What is the probability that it is grape or orange? 0.2 + 0.4 = 0.6

19c. What is the probability that it is not cherry? 1 - 0.3 = 0.7

20. Three candies are randomly selected from the bag.

20a. What is the probability that all three are grape? P(GGG) = 0.2(0.2)(0.2) = 0.008

20b. What is the probability that none are cherry? P(C'C'C') = 0.7(0.7)(0.7) = 0.343

20c. What is the probability that at least one is orange? 1 - P(No Orange) = 1 - 0.6(0.6)(0.6) = 1 - 0.216 = 0.784

20d. What is the probability that the last one is the first one that is lemon? P(L'L'L) = 0.9(0.9)(0.1) = 0.081

20e. What is the probability that all three are different flavors?

• P(GOC) = 0.2(0.4)(0.3) = 0.024
• P(GOL) = 0.2(0.4)(0.1) = 0.008
• P(GCL) = 0.2(0.3)(0.1) = 0.006
• P(OCL) = 0.4(0.3)(0.1) = 0.012

But there are 6 ways that each of those can occur, so add them together and multiply by 6.
6( 0.024 + 0.008 + 0.006 + 0.012 ) = 6 ( 0.050 ) = 0.300