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Zorluk: OrtaExperimental and Theoretical Probability

In a quality control experiment, a technician tested 200200 light bulbs and found that 3636 were defective. According to the manufacturer's specification, the theoretical probability of a bulb being defective is 320\frac{3}{20}. What is the difference between the theoretical probability and the experimental probability of selecting a non-defective light bulb?

  1. 3100\frac{3}{100}Cevap
  2. B
    67100\frac{67}{100}
  3. C
    33100\frac{33}{100}
  4. D
    17100\frac{17}{100}

Cevap

The correct answer is 3100\frac{3}{100} (or 0.030.03).
The experimental probability of obtaining a non-defective bulb is 1−36200=0.821 - \frac{36}{200} = 0.82. The theoretical probability of a non-defective bulb is 1−0.15=0.851 - 0.15 = 0.85. Subtracting the two gives 0.85−0.82=0.03=31000.85 - 0.82 = 0.03 = \frac{3}{100}.

Adım Adım Çözüm

1
Calculate the experimental probability of selecting a defective bulb
Pexp(defective)=36200=0.18\text{P}_{\text{exp}}(\text{defective}) = \frac{36}{200} = 0.18
Relative frequency is determined by dividing the observed frequency of defective bulbs by the total number of trials.
2
Find the experimental probability of selecting a non-defective bulb
Pexp(non-defective)=1−0.18=0.82\text{P}_{\text{exp}}(\text{non-defective}) = 1 - 0.18 = 0.82
The sum of the probabilities of an event and its complement equals 11.
3
Find the theoretical probability of selecting a non-defective bulb
Ptheo(non-defective)=1−320=1−0.15=0.85\text{P}_{\text{theo}}(\text{non-defective}) = 1 - \frac{3}{20} = 1 - 0.15 = 0.85
Subtracting the given theoretical probability of a defective bulb from 11 yields the theoretical probability of a non-defective bulb.
4
Calculate the difference between the two probabilities of selecting a non-defective bulb
0.85−0.82=0.03=31000.85 - 0.82 = 0.03 = \frac{3}{100}
Subtract the experimental probability from the theoretical probability.

Anahtar Kavram

Experimental probability measures relative frequency from observed outcomes, while theoretical probability is based on expected mathematical outcomes. Complementary probabilities fulfill P(E′)=1−P(E)P(E') = 1 - P(E).
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Experimental and Theoretical Probability

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