A dining facility manager is planning a special event and will randomly select one pie fro… — Armed Services Vocational Aptitude Battery (ASVAB)

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, $\frac{1}{3}$ of the pies are apple, $\frac{1}{4}$ are cherry, $\frac{1}{5}$ are blueberry, and the rest are pecan. What is the probability that the randomly selected pie is a pecan pie?

$\frac{13}{60}$ Answer
B
$\frac{13}{47}$
C
$\frac{47}{60}$
D
$\frac{3}{4}$

Answer

The probability of selecting a pecan pie is

\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

\frac{47}{60}

of the delivery. The remaining fraction,

\frac{13}{60}

, represents the probability of selecting a pecan pie.

Step-by-Step Solution

Find a common denominator to add the fractions of the known pie types (

\frac{1}{3}

\frac{1}{4}

, and

\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.

Convert each fraction to an equivalent fraction with a denominator of 60.

\frac{1}{3} = \frac{20}{60}

\frac{1}{4} = \frac{15}{60}

, and

\frac{1}{5} = \frac{12}{60}

This allows for straight addition of the numerators.

Add the equivalent fractions to find the total portion of apple, cherry, and blueberry pies.

\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.

Subtract this sum from 1 (the whole delivery) to find the probability of selecting a pecan pie.

1 - \frac{47}{60} = \frac{60}{60} - \frac{47}{60} = \frac{13}{60}

The sum of all possible mutually exclusive outcomes must equal 1.

Key Concept

Single Event Probability and Fraction Operations

Question

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Step-by-Step Solution

Key Concept

Common Mistakes

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