Engage NY Eureka Math 7th Grade Module 5 Lesson 6 Answer Key
Eureka Math Grade 7 Module 5 Lesson 6 Example Answer Key
Example 1: Two Nights of Games
Imagine that a family decides to play a game each night. They all agree to use a tetrahedral die (i.e., a four-sided pyramidal die where each of four possible outcomes is equally likely—see the image at the end of this lesson) each night to randomly determine if they will play a board game (B) or a card game (C). The tree diagram mapping the possible overall outcomes over two consecutive nights will be developed below.
To make a tree diagram, first present all possibilities for the first stage (in this case, Monday).
Then, from each branch of the first stage, attach all possibilities for the second stage (Tuesday).
Note: If the situation has more than two stages, this process would be repeated until all stages have been presented.
a. If BB represents two straight nights of board games, what does CB represent?
b. List the outcomes where exactly one board game is played over two days. How many outcomes were there?
Answer:
a. CB would represent a card game on the first night and a board game on the second night.
b. BC and CB—there are two outcomes.
Example 2: Two Nights of Games (with Probabilities)
In Example 1, each night’s outcome is the result of a chance experiment (rolling the tetrahedral die). Thus, there is a probability associated with each night’s outcome.
By multiplying the probabilities of the outcomes from each stage, we can obtain the probability for each “branch of the tree.” In this case, we can figure out the probability of each of our four outcomes: BB, BC, CB, and CC.
For this family, a card game will be played if the die lands showing a value of 1, and a board game will be played if the die lands showing a value of 2, 3, or 4. This makes the probability of a board game (B) on a given night 0.75.
a. The probabilities for two of the four outcomes are shown. Now, compute the probabilities for the two remaining outcomes.
b. What is the probability that there will be exactly one night of board games over the two nights?
Answer:
a. CB: (0.25)(0.75) = 0.1875
CC: (0.25)(0.25) = 0.0625
b. The two outcomes that contain exactly one night of board games are BC and CB. The probability of exactly one night of board games would be the sum of the probabilities of these outcomes (since the outcomes are disjoint). 0.1875 + 0.1875 = 0.375 (Note: Disjoint means two events that cannot both happen at once.)
Eureka Math Grade 7 Module 5 Lesson 6 Exercise Answer Key
Two friends meet at a grocery store and remark that a neighboring family just welcomed their second child. It turns out that both children in this family are girls, and they are not twins. One of the friends is curious about what the chances are of having 2 girls in a family’s first 2 births. Suppose that for each birth, the probability of a boy birth is 0.5 and the probability of a girl birth is also 0.5.
Exercise 1.
Draw a tree diagram demonstrating the four possible birth outcomes for a family with 2 children (no twins). Use the symbol B for the outcome of boy and G for the outcome of girl. Consider the first birth to be the first stage. (Refer to Example 1 if you need help getting started.)
Answer:
Exercise 2.
Write in the probabilities of each stage’s outcome to the tree diagram you developed above, and determine the probabilities for each of the 4 possible birth outcomes for a family with 2 children (no twins).
Answer:
In this case, since the probability of a boy is 0.5 and the probability of a girl is 0.5, each of the four outcomes will have a 0.25 probability of occurring because (0.5)(0.5) = 0.25.
Exercise 3.
What is the probability of a family having 2 girls in this situation? Is that greater than or less than the probability of having exactly 1 girl in 2 births?
Answer:
The probability of a family having 2 girls is 0.25. This is less than the probability of having exactly 1 girl in 2 births, which is 0.5 (the sum of the probabilities of BG and GB).
Eureka Math Grade 7 Module 5 Lesson 6 Problem Set Answer Key
Question 1.
Imagine that a family of three (Alice, Bill, and Chester) plays bingo at home every night. Each night, the chance that any one of the three players will win is \(\frac{1}{3}\).
a. Using A for Alice wins, B for Bill wins, and C for Chester wins, develop a tree diagram that shows the nine possible outcomes for two consecutive nights of play.
b. Is the probability that “Bill wins both nights” the same as the probability that “Alice wins the first night and Chester wins the second night”? Explain.
Answer:
a.
b. Yes. The probability of Bill winning both nights is \(\frac{1}{3}\) ∙ \(\frac{1}{3}\) = \(\frac{1}{9}\), which is the same as the probability of Alice winning the first night and Chester winning the second night (\(\frac{1}{3}\)∙\(\frac{1}{3}\) = \(\frac{1}{9}\)).
Question 2.
According to the Washington, D.C. Lottery’s website for its Cherry Blossom Doubler instant scratch game, the chance of winning a prize on a given ticket is about 17%. Imagine that a person stops at a convenience store on the way home from work every Monday and Tuesday to buy a scratcher ticket to play the game.
(Source: http://dclottery.com/games/scratchers/1223/cherry-blossom-doubler.aspx, accessed May 27, 2013)
a. Develop a tree diagram showing the four possible outcomes of playing over these two days. Call stage 1 “Monday,” and use the symbols W for a winning ticket and L for a non-winning ticket.
b. What is the chance that the player will not win on Monday but will win on Tuesday?
c. What is the chance that the player will win at least once during the two-day period?
Answer:
a.
b. LW outcome: (0.83)(0.17) = 0.1411
c. “Winning at least once” would include all outcomes except LL (which has a 0.6889 probability). The probabilities of these outcomes would sum to 0.3111.
Image of Tetrahedral Die
Source: http://commons.wikimedia.org/wiki/File:4-sided_dice_250.jpg
Photo by Fantasy, via Wikimedia Commons, is licensed under CC BY-SA 3.0,
http://creativecommons.org/licenses/by-sa/3.0/deed.en.
Eureka Math Grade 7 Module 5 Lesson 6 Exit Ticket Answer Key
In a laboratory experiment, two mice will be placed in a simple maze with one decision point where a mouse can turn either left (L) or right (R). When the first mouse arrives at the decision point, the direction it chooses is recorded. Then, the process is repeated for the second mouse.
Question 1.
Draw a tree diagram where the first stage represents the decision made by the first mouse and the second stage represents the decision made by the second mouse. Determine all four possible decision outcomes for the two mice.
Answer:
Question 2.
If the probability of turning left is 0.5 and the probability of turning right is 0.5 for each mouse, what is the probability that only one of the two mice will turn left?
Answer:
There are two outcomes that have exactly one mouse turning left: LR and RL. Each has a probability of 0.25, so the probability of only one of the two mice turning left is 0.25 + 0.25 = 0.5.
Question 3.
If the researchers add food in the simple maze such that the probability of each mouse turning left is now 0.7, what is the probability that only one of the two mice will turn left?
Answer:
As in Problem 2, there are two outcomes that have exactly one mouse turning left: LR and RL. However, with the adjustment made by the researcher, each of these outcomes now has a probability of 0.21. So now, the probability of only one of the two mice turning left is 0.21 + 0.21 = 0.42.
