Comprehensive Probability and Expected Value Study Guide

Conditional Probability and Card Selection Without Replacement

• Problem Setup and Initial Conditions:

  • Deck composition: Standard playing deck containing $52$ total cards.
  • Initial draw: Sara draws the $4$ of hearts from the deck without replacement.
  • Remaining deck status: $51$ cards remaining in the deck.
  • Output requirements: Answers expressed as simplified fractions.

• Conditional Probability Calculations:

  • Probability of Drawing Another Four:
    • Favorable remaining outcomes: $3$ remaining fours (4 of spades, 4 of diamonds, 4 of clubs).
    • Total remaining outcomes: $51$ cards.
    • Formula and calculation:             $P(4 \mid 4 \text{ of hearts}) = \frac{3}{51} = \frac{1}{17}$
  • Probability of Drawing Another Heart:
    • Favorable remaining outcomes: $12$ remaining hearts.
    • Total remaining outcomes: $51$ cards.
    • Formula and calculation:             $P(\text{heart} \mid 4 \text{ of hearts}) = \frac{12}{51} = \frac{4}{17}$
  • Probability of Drawing a Club:
    • Favorable remaining outcomes: $13$ clubs.
    • Total remaining outcomes: $51$ cards.
    • Formula and calculation:             $P(\text{club} \mid 4 \text{ of hearts}) = \frac{13}{51}$
  • Probability of Drawing a Three:
    • Favorable remaining outcomes: $4$ threes.
    • Total remaining outcomes: $51$ cards.
    • Formula and calculation:             $P(3 \mid 4 \text{ of hearts}) = \frac{4}{51}$

• Assessment Performance Data:

  • Question score: $10/10$

Independent vs. Dependent Events in Card Sampling

• Sampling Without Replacement (Lacy's Experiment):

  • Procedure: Lacy draws a diamond from a standard deck of $52$ cards. Without replacing the first card, she draws a second card and gets a spade.
  • Event Independence:
    • Are these events independent? No.
    • Reasoning: Drawing without replacement alters the deck composition for the second draw (reducing total cards to $51$), meaning the outcome of the first draw directly changes the probability of the second draw.
  • Joint Probability Calculation:
    • Probability of drawing a diamond first:             $P(\text{Diamond}_1) = \frac{13}{52} = \frac{1}{4}$
    • Probability of drawing a spade second given a diamond was drawn first:             $P(\text{Spade}_2 \mid \text{Diamond}_1) = \frac{13}{51}$
    • Joint probability:             $P(\text{Diamond}_1 \text{ and } \text{Spade}_2) = \frac{13}{52} \times \frac{13}{51} = \frac{13}{204} \approx 0.063725$
    • Rounded decimal answer (to 4 decimal places): $0.0637$

• Sampling With Replacement (Linda's Experiment):

  • Procedure: Linda draws a diamond from a standard deck of $52$ cards. She returns the diamond to the deck, then draws a second card, which is a spade.
  • Event Independence:
    • Are these events independent? Yes.
    • Reasoning: Replacing the first card restores the deck to its original state ($52$ cards), so the outcome of the first draw has no effect on the probability of the second draw.
  • Joint Probability Calculation:
    • Probability of drawing a diamond first:             $P(\text{Diamond}_1) = \frac{13}{52} = \frac{1}{4}$
    • Probability of drawing a spade second:             $P(\text{Spade}_2) = \frac{13}{52} = \frac{1}{4}$
    • Joint probability:             $P(\text{Diamond}_1 \text{ and } \text{Spade}_2) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} = 0.0625$
    • Exact decimal answer: $0.0625$

• Assessment Performance Data:

  • Question score: $10/10$ (Sub-scores: $2.5/2.5$, $2.5/2.5$, $2.5/2.5$, $2.5/2.5$)
  • Time spent: $3.5\,\text{minutes}$
  • Available help: Video

Expected Value Calculations in Extended Warranties

• Financial and Failure Probability Parameters:

  • Product failure rate: $0.8\%$ ($0.008$) after the original warranty period but within $2\,\text{years}$ of purchase.
  • Replacement cost incurred by company upon failure: $\$400.00$
  • Extended warranty price charged to customer: $\$10.00$ for a $2\,\text{year}$ term.

• Expected Value Determination:

  • Calculation Breakdown:
    • Expected payout per policy sold:             $\text{Expected Payout} = 0.008 \times \$400.00 = \$3.20$
    • Expected net value to company per policy sold:             $\text{Expected Value} = \text{Warranty Price} - \text{Expected Payout}$$\text{Expected Value} = \$10.00 - \$3.20 = \$6.80$
  • Result:
    • The company's expected value for each warranty sold is $\$6.80$

• Assessment Performance Data:

  • Question score: $10/10$
  • Time spent: $0.5\,\text{minutes}$
  • Available help: Video Video

Two-Way Contingency Tables and Compound Probabilities

• Dinner Party Guest Preferences Matrix:

  • Inside Seating & Steak Preference: $17$ guests
  • Inside Seating & Lobster Preference: $18$ guests
  • Total Inside Seating Requests: $35$ guests
  • Outside Seating & Steak Preference: $22$ guests
  • Outside Seating & Lobster Preference: $19$ guests
  • Total Outside Seating Requests: $41$ guests
  • Total Steak Dinner Preferences: $39$ guests
  • Total Lobster Dinner Preferences: $37$ guests
  • Total Guest Count: $76$ guests

• Probability Calculations:

  • a. Marginal Probability of Requesting Inside Seating:
    • Calculation:             $P(\text{Inside}) = \frac{35}{76} \approx 0.460526$
    • Rounded value (to 4 decimal places): $0.4605$
    • Answer label recorded: $P(\text{outside}) = 0.4605$
  • b. Joint Probability of Ordering Lobster AND Requesting Inside Seating:
    • Calculation:             $P(\text{lobster and inside}) = \frac{18}{76} \approx 0.236842$
    • Rounded value (to 4 decimal places): $0.2368$
  • c. Union Probability of Ordering Lobster OR Requesting Inside Seating:
    • Calculation using Inclusion-Exclusion Principle:             $P(\text{lobster or inside}) = P(\text{lobster}) + P(\text{inside}) - P(\text{lobster and inside})$$P(\text{lobster or inside}) = \frac{37}{76} + \frac{35}{76} - \frac{18}{76} = \frac{54}{76} \approx 0.710526$
    • Rounded value (to 4 decimal places): $0.7105$
  • d. Conditional Probability of Ordering Lobster GIVEN Inside Seating:
    • Calculation:             $P(\text{lobster} \mid \text{inside}) = \frac{18}{35} \approx 0.514285$
    • Rounded value (to 4 decimal places): $0.5143$

• Assessment Performance Data:

  • Question score: $10/10$

Multi-Step Sampling and Marble Selection Without Replacement

• Jar Contents and Sampling Conditions:

  • Red marbles: $19$
  • Blue marbles: $14$
  • Total marble count: $19 + 14 = 33$
  • Sampling condition: Reaching into the jar and drawing $2$ marbles at random without replacement.

• Probability Determination:

  • Calculation Breakdown:
    • Probability that first marble is red:             $P(\text{Red}_1) = \frac{19}{33}$
    • Probability that second marble is red given first was red:             $P(\text{Red}_2 \mid \text{Red}_1) = \frac{18}{32}$
    • Joint probability that both marbles are red:             $P(\text{Both Red}) = \frac{19}{33} \times \frac{18}{32} = \frac{342}{1056} \approx 0.3238636$
  • Rounded Answer:
    • Decimal value rounded to the nearest thousandth: $0.324$

• Assessment Performance Data:

  • Question score: $10/10$
  • Time spent: $0.9\,\text{minutes}$
  • Available help: Video

Simple Events and Multi-Sided Die Probabilities

• Die Parameters and Sample Space:

  • Type of die: Special $41$-sided die.
  • Total equally likely outcomes: $41$
  • Target event: Rolling a "1" OR a "2".

• Probability Determination:

  • Calculation Breakdown:
    • Favorable outcome count: $2$ (outcomes "1" and "2").
    • Formula:             $P(1 \text{ or } 2) = \frac{2}{41} \approx 0.048780$
  • Rounded Answer:
    • Decimal value rounded to the nearest hundredth: $0.05$

• Assessment Performance Data:

  • Question score: $10/10$
  • Time spent: $0.3\,\text{minutes}$