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On a street with 7 houses, 2 houses are blue. If a house from this street is selected at random,...

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

Source: Practice Test
Problem-Solving and Data Analysis
Probability and conditional probability
EASY
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Notes
Post a Query

On a street with \(\mathrm{7}\) houses, \(\mathrm{2}\) houses are blue. If a house from this street is selected at random, what is the probability of selecting a house that is blue?

A
\(\frac{1}{7}\)
B
\(\frac{2}{7}\)
C
\(\frac{5}{7}\)
D
\(\frac{7}{7}\)
Solution

1. TRANSLATE the problem information

  • Given information:
    • Total houses on the street: 7
    • Houses that are blue: 2
    • Selection method: random (each house equally likely to be chosen)
  • What we need to find: probability of selecting a blue house

2. INFER the approach

  • This is a basic probability problem asking for the chance of a favorable outcome
  • Use the fundamental probability formula: \(\mathrm{P(event)} = \frac{\mathrm{favorable\:outcomes}}{\mathrm{total\:outcomes}}\)
  • Favorable outcomes = blue houses = 2
  • Total outcomes = all houses = 7

3. Apply the probability formula

  • \(\mathrm{P(selecting\:blue\:house)} = \frac{\mathrm{Number\:of\:blue\:houses}}{\mathrm{Total\:number\:of\:houses}}\)
  • \(\mathrm{P(selecting\:blue\:house)} = \frac{2}{7}\)

Answer: B. 2/7


Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skills: Students correctly identify the numbers but mix up which goes in the numerator versus denominator.

Instead of thinking "2 blue houses out of 7 total," they focus on "5 houses are NOT blue" and calculate 5/7. This represents the probability of selecting a non-blue house rather than a blue house.

This leads them to select Choice C (5/7).

The Bottom Line:

This problem tests whether students can correctly identify the favorable outcomes (what they're looking for) versus the total possible outcomes, then set up the fraction properly. The key insight is that probability always asks "how many ways can the desired event happen" over "how many total ways are possible."

Answer Choices Explained
A
\(\frac{1}{7}\)
B
\(\frac{2}{7}\)
C
\(\frac{5}{7}\)
D
\(\frac{7}{7}\)
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