Two-Way Frequency Table & Joint Probability Solution Guide

Store Inventory Overview: Take Your Pick

  • Store Name: Take Your Pick
  • Total Instrument Inventory (NN): 6464 total instruments (guitars and pianos combined)
  • Instrument Categories:
    • Instrument Types: Guitars (GG) and Pianos (PP)
    • Power/Sound Types: Electric (EE) and Acoustic (AA)

Given Inventory Data

  • Total Electric Instruments (N(E)N(E)): 3636
  • Total Pianos (N(P)N(P)): 2929
  • Acoustic Pianos (N(P and A)N(P \text{ and } A)): 1212

Two-Way Frequency Table Construction

To find all missing quantities across the inventory categories, step-by-step arithmetic deductions are applied:

  • Total Acoustic Instruments (N(A)N(A)):N(A)=Total Instruments−N(E)=64−36=28N(A) = \text{Total Instruments} - N(E) = 64 - 36 = 28

  • Total Guitars (N(G)N(G)):N(G)=Total Instruments−N(P)=64−29=35N(G) = \text{Total Instruments} - N(P) = 64 - 29 = 35

  • Electric Pianos (N(P and E)N(P \text{ and } E)):N(P and E)=N(P)−N(P and A)=29−12=17N(P \text{ and } E) = N(P) - N(P \text{ and } A) = 29 - 12 = 17

  • Electric Guitars (N(G and E)N(G \text{ and } E)):N(G and E)=N(E)−N(P and E)=36−17=19N(G \text{ and } E) = N(E) - N(P \text{ and } E) = 36 - 17 = 19

  • Acoustic Guitars (N(G and A)N(G \text{ and } A)):N(G and A)=N(G)−N(G and E)=35−19=16N(G \text{ and } A) = N(G) - N(G \text{ and } E) = 35 - 19 = 16Check: N(A)−N(P and A)=28−12=16\text{Check: } N(A) - N(P \text{ and } A) = 28 - 12 = 16

Complete Contingency Table
Instrument TypeElectric (EE)Acoustic (AA)Total
Guitars (GG)191916163535
Pianos (PP)171712122929
Total363628286464

Probability Analysis: Piano and Electric Instrument

Target Event
  • Determining the probability that a randomly selected instrument from the store is both a piano and electric (P(Piano and Electric)P(\text{Piano and Electric})).
Mathematical Formula

P(Piano and Electric)=N(Piano and Electric)Total InventoryP(\text{Piano and Electric}) = \frac{N(\text{Piano and Electric})}{\text{Total Inventory}}

Calculation
  • Favorable Outcomes (N(P and E)N(P \text{ and } E)): 1717
  • Sample Space (NN): 6464

P(Piano and Electric)=1764P(\text{Piano and Electric}) = \frac{17}{64}

Equivalent Representations
  • Simplified Fraction: 1764\frac{17}{64}
  • Decimal Value: 0.2656250.265625
  • Percentage: 26.5625%26.5625\text{\%} or approximately 26.56%26.56\text{\%}

Related Marginal and Conditional Probabilities

  • Probability of Selecting Any Piano:P(Piano)=2964≈0.4531(45.31%)P(\text{Piano}) = \frac{29}{64} \thickapprox 0.4531 \quad (45.31\text{\%})

  • Probability of Selecting Any Electric Instrument:P(Electric)=3664=916=0.5625(56.25%)P(\text{Electric}) = \frac{36}{64} = \frac{9}{16} = 0.5625 \quad (56.25\text{\%})

  • Conditional Probability that a Piano is Electric (P(Electric  ∣  Piano)P(\text{Electric} \thickspace | \thickspace \text{Piano})):P(Electric  ∣  Piano)=N(P and E)N(P)=1729≈0.5862(58.62%)P(\text{Electric} \thickspace | \thickspace \text{Piano}) = \frac{N(P \text{ and } E)}{N(P)} = \frac{17}{29} \thickapprox 0.5862 \quad (58.62\text{\%})

  • Conditional Probability that an Electric Instrument is a Piano (P(Piano  ∣  Electric)P(\text{Piano} \thickspace | \thickspace \text{Electric})):P(Piano  ∣  Electric)=N(P and E)N(E)=1736≈0.4722(47.22%)P(\text{Piano} \thickspace | \thickspace \text{Electric}) = \frac{N(P \text{ and } E)}{N(E)} = \frac{17}{36} \thickapprox 0.4722 \quad (47.22\text{\%})