## Engage NY Eureka Math Precalculus Module 5 End of Module Assessment Answer Key

### Eureka Math Precalculus Module 5 End of Module Assessment Task Answer Key

Question 1.

An assembler of computer routers and modems uses parts from three sources. Company A supplies 60% of the parts, Company B supplies 30% of the parts, and Company C supplies the remaining 10% of the parts. From past experience, the assembler knows that 3% of the parts supplied by Company A are defective, 5% of the parts supplied by Company B are defective, and 8% of the parts supplied by Company C are defective. If a part is selected at random, what is the probability it is defective?

Answer:

P(defective) = P(defective|A)P(A) + P(defective|B)P(B) + P(defective|C)P(C)

= 0.6(0.03) + 0.3(0.05) + 0.1(0.08) = 0.041

Question 2.

In the game of roulette, a wheel with different slots is spun. A ball is placed in the wheel and bounces around until it settles in one of the 38 slots, all of equal size. There are 18 red slots, 18 black slots, and 2 green slots. Suppose it costs $1.00 to play one round of the game. A “color bet” allows you to pick either red or black; if the ball lands on your color, you win $2.00. If the ball lands on either of the other two colors, you do not win any money.

a. In this game, for each spin you win either $2.00 or $0.00. Determine the probabilities of winning each amount.

Answer:

P($2) = \(\frac{18}{38}\) ≈ 0.4737

P($0) = \(\frac{20}{38}\) ≈ 0.5263

b. Calculate the expected winnings in one spin of the wheel.

Answer:

Expected winnings = 2(18/38) + 0(20/38) = \(\frac{36}{38}\) ≈ 0.9474

The expected winnings in one spin of the wheel is $0.95.

c. Are your expected winnings in one spin of the wheel larger or smaller than the $1.00 bet?

Answer:

This is smaller than the $1.00 bet (about 5 cents).

d. A casino’s profit is equal to the amount of bets minus the amount of winnings. For each spin of the roulette wheel, the expected profit is $1.00 minus your expected winnings. What is the casino’s expected profit? Explain why the game of roulette is still attractive to a professional casino even though the expected profit to the casino is such a small amount on each spin.

Answer:

The casino’s expected winnings are about 5 cents per spin. The law of large numbers states that if they can get enough people to play, the average winnings will be very close to the expected winnings. So, the casino will be pretty sure of being close to the expected winnings. Even a small amount of expected winnings will add up over a large number of spins.

e. The slots are also numbered 0, 00, 1–36. If you bet on a number and win, you win $36.00. Which bet is better for you: a number bet or a color bet? Explain your decision.

Answer:

Expected winnings = 36(\(\frac{1}{38}\)) = \(\frac{36}{38}\) .

The expected winnings are the same for the two bets. (So, you could argue that there isn’t a difference, or you could argue in terms of the thrill versus security of the bets.)

f. Suppose you plan to construct a wheel with 12 slots: 5 red, 5 black, and 2 green. You plan to pay $5.00 in winnings if someone picks a color (red or black) and the ball lands on that color. How much should you charge someone to play (the bet amount) so that you have created a fair game?

Answer:

Expected winnings = 5(\(\frac{5}{12}\)) = \(\frac{25}{12}\). So, you should charge approximately $2.08 to break even. You should also accept $2.09 to cover the rounding from $2.083.

Question 3.

In New York, the Mega Ball jackpot lottery asks you to pick 5 numbers (integers) from 1 to 59 (the “upper section”) and then pick a Mega Ball number from 1 to 35 (the “lower section”). You win $10,000 if you match exactly 4 of the 5 numbers from the upper section and match the Mega Ball number from the lower section. The winning number or numbers for each section are chosen at random without replacement. Determine the probability of correctly choosing exactly 4 of the winning numbers and the Mega Ball number.

Answer:

P(4 winning numbers and 1 Mega Ball) = (\(\frac{1}{{ }_{59} C_{5}}\)) (\(\frac{1}{{ }_{35} C_{1}}\)).

The probability of correctly choosing exactly 4 of the winning numbers and the Mega Ball number is 0.00000000571 (5.71 x10-9).

Question 4.

A blood bank is screening a large population of donations for a particular virus. Suppose 5% of the blood donations contain the virus. Suppose you randomly select 10 bags of donated blood to screen. You decide to take a small amount from each of the bags and pool these all together into one sample. If that sample shows signs of the virus, you then test all of the original 10 bags individually. If the combined sample does not contain the virus, then you are done after just the one test.

a. Determine the expected number of tests to screen 10 bags of donated blood using this strategy.

Answer:

If none of the blood donations have the virus, then you only have to conduct one test. The probability that none of the (independent) people have the virus is

(1 – 0.05)^{10} ≈ 0.5987.

So, the expected number of tests is 1(0.5987) + 11(1 – 0.5987) = 5.01.

b. Does this appear to be an effective strategy? Explain how you know.

Answer:

This expected value in part (a) is much less than 10, so this seems to be an effective strategy. In the long run, you will perform fewer tests.

Question 5.

Twenty-five sixth-grade students entered a math contest consisting of 20 questions. The student who answered the greatest number of questions correctly will receive a graphing calculator. The rules of the contest state that if two or more students tie for the greatest number of correct answers, one of these students will be chosen to receive the calculator.

No student answered all 20 questions correctly, but four students (Allan, Beth, Carlos, and Denesha) each answered 19 questions correctly.

What would be a fair way to use two coins (a dime and a nickel) to decide which student should get the calculator? Explain what makes your method fair.

Answer:

Answers will vary. One possible response is to toss the coins, and if both coins land heads up (HH), Allan gets the calculator. If the dime lands heads up and the nickel lands tails up (HT), Beth gets the calculator. If the dime lands tails up and the nickel lands heads up (TH), Carlos gets the calculator. If both coins land tails up (TT), Denesha gets the calculator. This method is fair because each of the four students has the same chance (a probability of \(\frac{1}{4}\) of getting the calculator.

Question 6.

A cell phone company offers cell phone insurance for $7.00 a month. If your phone breaks and you submit a claim, you must first pay a $200.00 deductible before the cell phone company pays anything. Suppose the replacement cost for a phone is $650.00. This means if you break your phone and have insurance, you have to pay only $200.00 toward the replacement cost. This plan has a limit of two replacements; if you break your phone more than twice in one year, you pay for the full replacement cost for the additional replacements.

Suppose that within one year, there is a 48% chance that you do not break your phone, a 36% chance that you break it once, a 12% chance that you break it twice, a 3% chance that you break it three times, and a 1% chance that you break it four times.

a. Calculate the expected one-year cost of this insurance plan based on the monthly cost and the expected repair costs.

Answer:

If you do not break your phone, then you pay $7 × 12 = $84.

If you break your phone once, then you pay $200 + $84 = $284.

If you break your phone twice, then you pay $400 + $84 = $484.

If you break your phone three times, then you pay $484 + $650 = $1134.

If you break your phone four times, then you pay $1134 + $650 = $1784.

The probability distribution of costs is as follows:

Therefore, your expected costs are

0.48($84) + 0.36($284) + 0.12($484) + 0.03($1134) + 0.01($1784) = $252.50

b. Determine your expected replacement costs if you do not purchase insurance.

Answer:

0.48($0) + 0.36($650) + 0.12($1300) + 0.03($1950) + 0.01($2600) = $474.50

c. Does this insurance plan seem to be a good deal? Explain why or why not.

Answer:

Yes. The long-run average costs are considerably less with the insurance plan than without it.