Applied Probability Trivia Game | Math

Quiz Answer Key and Fun Facts

1. A duelist playing the "Yu Gi Oh" trading card game wants to win using the Exodia cards. These are a set of five different cards, each containing a piece of Exodia's body. In the game, each player draws a starting hand of five cards. Assuming the player has a deck of 40 cards which contains a complete Exodia set and where all the cards are different (i.e. contains no duplicate cards), what are the odds that their starting hand of five cards will be the complete Exodia set?

Answer: 1 / 658,008

Since all 40 cards are different and order does not matter, we must choose 5 cards out of 40. We use the formula 40! / (35! * 5!) to give us the number of combinations. (n! means n*(n-1)*(n-2)*...*3*2*1.) This gives (40*39*38*37*36) / (5*4*3*2*1), which gives 658,008 combinations.

There is only one combination that consists of all five pieces of Exodia, so the odds of starting off with it are 1 / 658,008.

2. A gambler in a casino is playing Craps. In this game, he rolls two fair dice at the same time and adds the two faces that come up. If his first roll is a seven or eleven, he wins. If his first roll is a two, three, or twelve, he loses. If it is anything else, he keeps rolling until he gets either a seven (in which case he loses) or whatever number he rolled (in which case he wins). What are the odds the gambler will win his next game of Craps?

Answer: 244 / 495

The probability of rolling a seven on the first roll is 6 / 36 = 1 / 6. Similarly, an eleven will be rolled with a probability of 2 / 36 = 1 / 18. Next, a four will be rolled with a probability of 3 / 36 = 1 / 12. Next, we want something other than a four or a seven to be rolled, then we want a four to be rolled.

This is [(1 - (1 / 6) - (1 / 12))^n] * (1 / 12), where n can be anything from zero to infinity. Summing the amounts for all values of n using the formula (first term / (1 - ratio)), we get 1 / 3, which we multiply by the original probability of rolling a four on the first roll to get 1 / 36 as the probability of winning a game of Craps with a first roll of four. By symmetry, this is also the probability of winning with a ten as your first roll. Similarly, we get the odds of winning with a five (symmetric to nine) and a six (symmetric to eight) to be 2 / 45 and 25 / 396 respectively. Adding up the probabilities of all the different ways to win, we get 244 / 495.

3. To pick a winning 4 digit lottery number, a lottery commissioner has four urns. Each urn contains ten balls. The balls in each urn are numbered zero through nine. One ball is picked out of each urn to make up one digit of the winning number. What are the odds that the winning number consists only of nines or zeros?

Answer: 1 / 625

We want each number to be a zero or nine, and there is a (2 / 10) = (1 / 5) of this happening for each number. Therefore, there is a (1 / 5)^4 = 1 / 625 chance of the winning number containing only zeros and nines.

4. A bus is guaranteed to arrive at the bus stop between 8:00 am and 8:15 am. It is no more likely to arrive at any time than any other time in this interval (i.e. the probability density function is uniform). What is the probability that the bus will arrive at 8:05 am?

Answer: 0

Time is continuous, so there can not be a specific time at which the bus arrives. In other words, there are an infinite amount of times between 8:00 am and 8:15 am, so the odds of the bus arriving at any one time are 1 / infinity = 0. Or if you'd prefer, the odds of the bus arriving between 8:04 am and 8:06 am are 2 / 15, since the bus is equally likely to arrive at towards the beginning of the interval [8:00 am , 8:15 am] and we are looking at a subinterval of 2 minutes out of 15. If we keep shortening the subinterval, the probability that the bus will arrive within it keep getting smaller. We can keep shrinking the size of the subinterval arbitrarily, so we can keep making the probability smaller.

5. When flipping a fair coin, we would obviously expect to get 5 heads out of 10 tosses, but what are the odds of this actually happening?

Answer: 63 / 256

The odds of getting a head (and subsequently a tail) on any particular toss is 1 /2. Therefore the odds of getting any particular arrangement of heads and tails is [(1 / 2)^5 * (1 / 2)^5] = (1 / 2)^10 = 1 / 1,024. However there are 10! / (5! * 5!) ways to arrange five heads and five tails.

This is (10 * 9 * 8 * 7 * 6 / 5 / 4 / 3 / 2 / 1) = 252 arrangements. The odds we are looking for thus are 252 / 1,024 = 63 / 256.

6. An insurance company determines that the probabilities of a driver in their 20s, 30s, 40s, 50s, and 60s getting into a crash at least once within these respective decades are 0.9, 0.6, 0.5, 0.4, and 0.75 respectively. A random driver signs up for coverage on their 20th birthday. Ignoring any other factors, what are the chances that they will get into their first accident in their 60s?

Answer: 0.009

The driver must avoid accidents during their 20s, 30s, 40s, and 50s, with probabilities (1 - 0.9), (1 - 0.6), (1 - 0.5), and (1 - 0.4) respectively. They must then get into a crash during their 60s with probability 0.75. Multiplying these together gives (0.1 * 0.4 * 0.5 * 0.6 * 0.75) = 0.009. The calculation assumes that each decade's risk factor is independent of the others.

7. The lifetime of a copy machine from the time of purchase is described by the probability density function f(x)=.5*e^(-.5*x). A warranty refunds the purchase price of the machine if it dies within one year after purchase. What is the probability that this warranty will be paid out for a randomly selected machine?

Answer: 1 - e^(-1/2)

We want the probability that a random machine will die between zero and one years after purchase. To do this, we integrate the probability density function from 0 to 1. This means evaluating -e^(-.5*x) from 0 to 1, which is -e^(-.5) + 1, or 1 - e^(-1/2).

8. An insurance company determines that the probability of the Boston Red Sox winning the World Series to be (3/32). Jordan's Furniture is running a promotion that will allow customers to keep any furniture they purchased between March 8 and April 16 for free. To cover against potential losses, Jordan's wants to take out an insurance policy that will pay them $20,000,000 if the Red Sox win the World Series. How much should the insurance company charge for this policy to cancel out their expected payout?

Answer: $1,875,000

Their expected payout is the amount they would have to pay multiplied by the probability of this happening, in this case $20,000,000 * (3/32) = $1,875,000. The insurance company would have to charge Jordan's Furniture this amount in order to recoup their expected losses.

9. In poker (played with a regular 52 card deck), a royal flush is when you have a ten, jack, queen, king, and ace all of the same suit. If a man is playing five card stud poker (where you are dealt five cards and can not draw more), what is the probability that he will be dealt a royal flush?

Answer: 1 / 649,740

There are 52! / (47! * 5*) = (52 * 51 * 50 * 49 * 48 / (5 * 4 * 3 * 2 * 1)) = 2,598,960 possible hands of five cards. Of these four are royal flushes (the ten, jack, queen, king, and ace of hearts, diamonds, clubs, and spades). Therefore the probability of getting a royal flush is 4 / 2,598,960 = 1 / 649,740.

10. The probability density function of the number of seconds it takes for the eyes of a random patient involved in a study to adjust to the light in a room is f(x) = 3 * x^2 from 0 to 1. What is the probability that a randomly selected patient's eyes will adjust to a room's lighting in between (1 / 2) and (3 / 4) seconds?

Answer: 0.296875

We must integrate the density function f(x) = 3 * x^2 from (1 / 2) to (3 / 4). This means evaluating x^3 from (1 / 2) to (3 / 4), which is (3 / 4)^3 - (1 / 2)^3, which is (27 / 64) - (1 / 8), which is (19 / 64), which is 0.296875.

Source: Author jrrymaury

This quiz was reviewed by FunTrivia editor crisw before going online.
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Applied Probability Trivia Quiz

In this quiz, I feature probability problems that I have related to real-life situations to make the questions applicable to daily life. Both finite and continuous distributions are reviewed. Some integration is needed for computing continuous examples.

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