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Zorluk: OrtaSingle Event Probability

A dining facility manager is planning a special event and will randomly select one pie from a large frozen delivery. According to the shipping manifest, 13\frac{1}{3} of the pies are apple, 14\frac{1}{4} are cherry, 15\frac{1}{5} are blueberry, and the rest are pecan. What is the probability that the randomly selected pie is a pecan pie?

  1. 1360\frac{13}{60}Cevap
  2. B
    1347\frac{13}{47}
  3. C
    4760\frac{47}{60}
  4. D
    34\frac{3}{4}

Cevap

The probability of selecting a pecan pie is 1360\frac{13}{60}.
The sum of the probabilities of all possible outcomes in this scenario must equal 1. By finding a common denominator (60), we determine that apple, cherry, and blueberry pies account for 4760\frac{47}{60} of the delivery. The remaining fraction, 1360\frac{13}{60}, represents the probability of selecting a pecan pie.

Adım Adım Çözüm

1
Find a common denominator to add the fractions of the known pie types (13\frac{1}{3}, 14\frac{1}{4}, and 15\frac{1}{5}).
The least common multiple of 3, 4, and 5 is 60.
Fractions must have the same denominator before they can be added.
2
Convert each fraction to an equivalent fraction with a denominator of 60.
13=2060\frac{1}{3} = \frac{20}{60}, 14=1560\frac{1}{4} = \frac{15}{60}, and 15=1260\frac{1}{5} = \frac{12}{60}.
This allows for straight addition of the numerators.
3
Add the equivalent fractions to find the total portion of apple, cherry, and blueberry pies.
2060+1560+1260=4760\frac{20}{60} + \frac{15}{60} + \frac{12}{60} = \frac{47}{60}.
This sum represents all the pies in the delivery that are NOT pecan.
4
Subtract this sum from 1 (the whole delivery) to find the probability of selecting a pecan pie.
1−4760=6060−4760=13601 - \frac{47}{60} = \frac{60}{60} - \frac{47}{60} = \frac{13}{60}.
The sum of all possible mutually exclusive outcomes must equal 1.

Anahtar Kavram

Single Event Probability and Fraction Operations
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Konular

Single Event Probability

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