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Zorluk: OrtaSingle Event Probability

A motor pool inventory shows that 14\frac{1}{4} of the vehicles are heavy trucks, 13\frac{1}{3} are light tactical vehicles, and the remaining vehicles are armored personnel carriers. If a vehicle is randomly selected for an inspection, what is the probability that it is an armored personnel carrier?

  1. A
    712\frac{7}{12}
  2. B
    27\frac{2}{7}
  3. 512\frac{5}{12}Cevap
  4. D
    57\frac{5}{7}

Cevap

512\frac{5}{12}
The total probability of all mutually exclusive events must sum to 1. First, find the combined probability of heavy trucks and light tactical vehicles by using a common denominator of 12: 14\frac{1}{4} becomes 312\frac{3}{12} and 13\frac{1}{3} becomes 412\frac{4}{12}. Their sum is 712\frac{7}{12}. Since the remaining vehicles are armored personnel carriers, subtract this sum from the total probability of 1 (or 1212\frac{12}{12}). The result is 512\frac{5}{12}.

Adım Adım Çözüm

1
Find the combined probability of selecting a heavy truck or a light tactical vehicle.
14+13\frac{1}{4} + \frac{1}{3}
To determine the portion of the fleet that is not armored personnel carriers, the given fractions must be added.
2
Add the fractions by finding a common denominator.
312+412=712\frac{3}{12} + \frac{4}{12} = \frac{7}{12}
Fractions must have the same denominator (12) before their numerators can be added.
3
Subtract the combined fraction from 1 (the total probability).
1−712=1212−712=5121 - \frac{7}{12} = \frac{12}{12} - \frac{7}{12} = \frac{5}{12}
The sum of probabilities for all possible events must equal exactly 1. The remainder represents the armored personnel carriers.

Anahtar Kavram

Single Event Probability from Complementary Events
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