A Standard Deck Has 52 Playing Cards

A Standard Deck Has 52 Playing Cards There Are

Questions: (20 points) A standard deck has 52 playing cards. There are thirteen ranks, ranging from 2 up to 10, then Jack, Queen, King and Ace. Each rank has exactly one card each of four different suits: Spades ♠, Hearts ♥, Diamonds ♦, and Clubs ♣. a. Suppose you select two cards at random from the deck. What is the probability that you have a pair worse than a pair of Jacks? (The cards 2-10 are all worse than Jacks) The game of Pinochle (pronounced “pea-knuckle”) is played with a non-standard deck of cards. The deck only contains 9s, 10s, Jacks, Queens, Kings, and Aces. It contains two copies of each of these cards in each suit (so there are 2 copies of the 9♥, two copies of the K♠, and so on). An important card combination is the Pinochle, which consists of the Q♦ and the J♣. Suppose two cards are drawn at random from a Pinochle deck. What is the probability they form a Pinochle? (Bonus 10 points) The most popular card game in the world (at least for tournament competition) is Contract Bridge. In the game of Bridge, each of 4 players receives a hand of 13 cards from the standard 52 card deck. (Therefore the whole deck is dealt out to the players). Bridge is a very complicated game, but one important way in which players decide how to play their hand is by counting the number of “High Card Points”. Each ace is worth 4 points, each king 3, queen 2, and jack 1 point. Let X be the number of high card points a player has in their hand of 13 cards. Find E[X]. (15 points) Using historical records, the manager of a construction site has determined that the number of minor injuries suffered by workers per day has the following distribution : Find the following probabilities: a) The number of injuries is in the interval [3,6]. b) At least 1 injury occurs. c) Fewer than 4 injuries occur. 3. (20 points) You are a human resources professional. Your company is hiring for two different jobs. The first job is in sales, the second job is in finance. You have a list of candidates, who may be worth interviewing for one of the jobs, both of the jobs, or neither. The probability that a candidate is suitable for the sales job is .45, the probability that a candidate is suitable for the finance job is .35, and the probability of being suitable for both is .15. Answer the following questions: a. What is the probability that a candidate is unhirable? b. What is the probability that a candidate is worth interviewing for at least one job? c. What is the probability that a candidate is worth interviewing for the sales job, but not the finance job? d. Let X be the number of jobs for which a candidate is worth interviewing. Find E[X]. E P(E) .25 .25 .15 .10 .10 .08 . points) Actuaries are insurance professionals. Becoming an actuary requires passing a series of 9 exams. These exams are offered twice a year, and people traditionally take them while employed as junior actuaries. Therefore, actuarial jobs normally give workers paid study time. This can be a problem because study time can be used for activities other than studying, and studying is no guarantee one will pass a given exam. (It normally takes about 7 years to pass all 9 exams, which implies a failure rate of around 33% for people who actually pass all 9). Suppose 50% of all actuarial students study hard, and have an 80% chance of passing a given exam. The other 50% cram for the exam at the last minute, and have a 40% chance of passing the exam. You are the manager of some junior actuaries. One of them just took an exam and failed. What is the probability that they studied hard for the exam? (15 points) A marketing manager for a department store used a survey to determine how a sample of customers learned about the store’s products, as well as the gender of the customer. The joint probabilities are reported in the following table: a) Given that a customer is female, what is the probability she learned about the store online? b) Given that the customer learned about the store via a newspaper, what is the probability they are male? c) What is the probability a customer is female? d) What is the probability a customer is both male and learned about the store on TV? e) Let E be the event {Online} and F be the event {Male}. Are E and F independent? (15 points) Suppose you are attending a party, which happens to be rather boring. You are talking with various people about their birthdays. a. Suppose there are 8 people attending the party. What is the probability that everyone was born in a different month? Assume that all partygoers will tell the truth about their birth month, and that people are equally likely to be born in any month. b. Assume now that 10 people are attending the party. What is the probability that everyone was born on a different day of the month (For example, my birthday is January 18th. My day of the month would be the 18th.) For convenience, assume that all months have 30 days, and every day is equally likely to be someone’s birthday.

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Introduction

The set of questions presented covers a broad spectrum of probability and statistics topics, including card probability calculations, expected value computations, probability distribution analysis, and applied probability scenarios involving real-world contexts such as workplace accidents, hiring decisions, actuarial exams, consumer behavior surveys, and birthday probabilities. This comprehensive discussion aims to explore each problem systematically, applying fundamental principles of probability theory, combinatorics, and statistical analysis to arrive at precise solutions supported by relevant theoretical frameworks.

Question 1: Probability of a Pair Worse Than Jacks in a Standard Deck

In a standard deck with 52 cards, there are 13 ranks and 4 suits per rank. To calculate the probability of selecting two cards that form a pair worse than Jacks—that is, two cards of the same rank, with the rank less than Jack (i.e., 2 through 10)—we first identify the favorable and total outcomes.

Favorable outcomes involve choosing any of the 9 possible ranks (2-10); from each rank, selecting 2 suits out of 4 (since each suit has exactly one card per rank). The number of ways to choose 2 suits from 4 is C(4,2) = 6. Thus, total favorable pairs per rank are 6.

Since there are 9 such ranks, the total favorable pairs are 9 x 6 = 54.

The total number of ways to select any 2 cards from 52 is C(52,2) = 1326.

Therefore, the probability is P = 54 / 1326 ≈ 0.0407.

Question 2: Probability of forming a Pinochle from a Pinochle Deck

The Pinochle deck comprises 48 cards: 2 copies each of 9, 10, J, Q, K, and A in four suits. The total number of possible pairs is C(48, 2) = 1128. The Pinochle combination consists of a Jack of Clubs (J♣) and a Queen of Diamonds (Q♦). Since there are exactly 2 copies of each card, there are 2 copies of J♣ and 2 copies of Q♦, making 4 possible pairs (each Jack of Clubs with each Queen of Diamonds).

Thus, the probability is 4 / 1128 ≈ 0.00355.

Question 3: Expected High Card Points in a Bridge Hand

In Bridge, each player receives 13 cards from a standard deck. High card points (HCP) are assigned as follows: Ace = 4, King = 3, Queen = 2, Jack = 1. Since the deck is well-shuffled, each card is equally likely to appear in a player's hand.

Calculating E[X] involves summing the expected points for each rank across the 13 cards.

The probability that a specific card of rank r is in the hand is 13 / 52 = 1/4, due to symmetry.

Each rank has 4 cards; hence, the expected number of cards of a given rank in the hand is 4 x (13/52) = 1.

Therefore, the expected points contributed by each rank are:

  • For Ace: 4 points x expected count of Ace (1) = 4
  • Similarly, for King: 3 x 1 = 3
  • Queen: 2 x 1 = 2
  • Jack: 1 x 1 = 1

Summing these for all four ranks gives:

Expected total high card points: (4 + 3 + 2 + 1) = 10.

Thus, E[X] = 10 points.

Question 4: Distribution of Minor Injuries at a Construction Site

Assuming the distribution of minor injuries per day is known, the probability calculations depend on using the probability mass function or frequency distribution information provided, which is not specified explicitly here. Typically, such questions involve summing probabilities over the relevant intervals, such as:

  • a) probability injuries are in [3,6],
  • b) probability at least 1 injury occurs,
  • c) probability fewer than 4 injuries occur.

Without explicit distribution data, solutions involve summing the probabilities from the given pmf or data entries over the specified ranges.

Question 5: Hiring Probabilities for Candidates

Let S be the event candidate is suitable for sales (probability 0.45), F for finance (probability 0.35), and B for both (probability 0.15).

  • a) Probability candidate is unhirable: P(neither S nor F) = 1 - P(suitable for at least one). Since P(S) + P(F) - P(B) = 0.45 + 0.35 - 0.15 = 0.65, then unhirable = 1 - 0.65 = 0.35.
  • b) Probability candidate worth interviewing for at least one job: 0.65.
  • c) Worth interviewing for sales but not finance: P(S and not F) = P(S) - P(B) = 0.45 - 0.15 = 0.30.
  • d) The expected value E[X] of the number of jobs a candidate is suitable for: Since X can be 0, 1, or 2, with probabilities P(0) = P(neither), P(1) = P(S only) + P(F only), P(2) = P(both).

    P(S only) = P(S) - P(B) = 0.30; P(F only) = P(F) - P(B) = 0.20; P(both) = 0.15; P(none) = 0.35.

    Therefore, E[X] = 00.35 + 1(0.30 + 0.20) + 2*0.15 = 0 + 0.50 + 0.30 = 0.80.

Question 6: Probability of Passing All 9 Exams and Studying Hard

Applying Bayesian reasoning, if a student studies hard (probability 0.5) with an 80% chance of passing each exam, and crams (probability 0.5) with a 40% chance, then:

Given the student failed an exam, the probability that they studied hard is given by:

P(studied hard | failed) = P(failed | studied hard) * P(studied hard) / P(failed).

P(failed | studied hard) = 1 - P(passed | studied hard) = 1 - 0.8 = 0.2;

P(failed | cramming) = 1 - 0.4 = 0.6.

P(failed) = P(failed | hard) P(hard) + P(failed | cramming) P(cramming) = 0.20.5 + 0.60.5 = 0.1 + 0.3 = 0.4.

Thus, P(studied hard | failed) = (0.2*0.5) / 0.4 = 0.1 / 0.4 = 0.25.

Question 7: Customer Survey Probabilities and Independence

The joint probabilities enable calculating marginal probabilities and testing independence.

  • a) P(Female and learned online) = P(Female ∩ Online) / P(Female) = (joint probability) / (sum over online in females). Without the specific table, the process involves dividing the joint probability for females who learned online by the total probability of females.
  • b) P(Male | Newspaper) = P(Male ∩ Newspaper) / P(Newspaper). Similar approach.
  • c) Probability a customer is female: sum the probabilities over all events where customer is female divided by total.
  • d) P(Both male and learned on TV): specific joint probability as per the table.
  • e) Independence of E and F involves verifying whether P(E ∩ F) = P(E) * P(F).

Question 8: Probability of Unique Birth Months and Days

a) For 8 people, the probability all are born in different months: P = (12/12) (11/12) (10/12) ... (5/12) for the second through eighth person, assuming each person is equally likely to be born in any month, with no repeats. The calculation: P = 12/12 11/12 10/12 9/12 8/12 7/12 6/12 * 5/12.

b) For 10 people and 30 days in each month, probability everyone is born on different days: P = 30/30 29/30 28/30 ... (30-9)/30.

This calculations reflect permutations without replacement, assuming uniform distributions.

Conclusion

The problems addressed emphasize core probability concepts such as combinatorial calculations, expectation, joint and marginal probabilities, and the use of Bayes' theorem. Application of these principles to real-world scenarios demonstrates their utility and importance in decision-making, risk assessment, and statistical analysis.

References

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