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Zorluk: OrtaTwo-Way Tables and Probability

A municipal parking garage tracked the charging times of 200200 electric vehicles over a one-week period. The vehicles were classified by type—Battery Electric Vehicles (BEVs) or Plug-in Hybrid Electric Vehicles (PHEVs)—and by the time of day they primarily charged: Daytime (6:00 AM to 6:00 PM) or Nighttime (6:00 PM to 6:00 AM). The results are summarized in the table below.

Vehicle TypeDaytime ChargingNighttime ChargingTotal
BEV45457575120120
PHEV555525258080
Total100100100100200200

If one of these electric vehicles is selected at random, and it is found to have charged during the nighttime, what is the probability that the vehicle is a PHEV?

  1. A
    18\frac{1}{8}
  2. 14\frac{1}{4}Cevap
  3. C
    516\frac{5}{16}
  4. D
    38\frac{3}{8}

Cevap

14\frac{1}{4}
The correct answer is the simplified fraction representing the number of nighttime-charging PHEVs (2525) divided by the total number of nighttime-charging vehicles (100100), which is 25100=14\frac{25}{100} = \frac{1}{4}.

Adım Adım Çözüm

1
Identify the total number of vehicles in the restricted sample space.
The total number of vehicles that charged during the nighttime is 100100.
The question specifies that the vehicle 'is found to have charged during the nighttime', which limits the denominator to the 'Nighttime Charging' column total.
2
Determine the number of favorable outcomes within this restricted sample space.
The number of PHEVs that charged during the nighttime is 2525.
We need to find the count in the intersection of the 'PHEV' row and the 'Nighttime Charging' column.
3
Calculate the conditional probability as a fraction.
The probability is 25100=14\frac{25}{100} = \frac{1}{4}.
Dividing the number of favorable vehicles (2525) by the conditional total (100100) yields the final probability, which simplifies to 14\frac{1}{4}.

Anahtar Kavram

Conditional probability calculations from a two-way table involve restricting the sample space to a specific row or column total rather than the grand total.
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