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A playlist contains 15 distinct songs. The playlist is shuffled uniformly at random, and the first song is played. What...

GMAT Problem-Solving and Data Analysis : (PS_DA) Questions

Source: Prism
Problem-Solving and Data Analysis
Probability and conditional probability
EASY
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A playlist contains 15 distinct songs. The playlist is shuffled uniformly at random, and the first song is played. What is the probability that a particular song, Song T, is the first song played?

  1. \(\frac{1}{15}\)
  2. \(\frac{1}{14}\)
  3. \(\frac{2}{15}\)
  4. \(\frac{1}{16}\)
A
\(\frac{1}{15}\)
B
\(\frac{1}{14}\)
C
\(\frac{2}{15}\)
D
\(\frac{1}{16}\)
Solution

1. TRANSLATE the problem information

  • Given information:
    • 15 distinct songs in playlist
    • Playlist is shuffled uniformly at random
    • We want probability that Song T is played first
  • What this tells us: We need to find \(\mathrm{P(Song\;T\;is\;in\;first\;position)}\)

2. INFER the probability structure

  • Key insight: 'Uniformly at random' means every possible arrangement of the 15 songs is equally likely
  • Since all arrangements are equally likely, each individual song has the same chance of ending up in the first position
  • This becomes a simple equal-outcomes probability problem

3. INFER the calculation approach

  • Total possible songs that could be first: 15 (any of the 15 songs)
  • Favorable outcomes: 1 (only Song T satisfies our condition)
  • Since each song is equally likely to be first: \(\mathrm{P(Song\;T\;first)} = \frac{1}{15}\)

Answer: A. \(\frac{1}{15}\)




Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students misinterpret what happens after the first song is selected, thinking there are only 14 songs left to consider for the probability calculation.

They incorrectly reason: 'After Song T is selected, there are 14 remaining songs, so the probability should be 1/14.' This confuses the probability of Song T being first with some other conditional probability scenario.

This may lead them to select Choice B (\(\frac{1}{14}\)).

The Bottom Line:

This problem tests whether students can recognize a basic equal-probability situation. The key insight is that when distinct objects are arranged randomly, each object has the same probability of occupying any specific position - in this case, 1 out of 15 for the first position.

Answer Choices Explained
A
\(\frac{1}{15}\)
B
\(\frac{1}{14}\)
C
\(\frac{2}{15}\)
D
\(\frac{1}{16}\)
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