Home / Digital SAT Math Practice Questions – Medium : Probability and conditional probability

Digital SAT Math Practice Questions – Medium : Probability and conditional probability

SAT MAth Practice questions – all topics

  • Problem-solving and Data Analysis Weightage: 15%  Questions: 5-7
    • Ratios, rates, proportional relationships, and units
    • Percentages
    • One-variable data: distributions and measures of centre and spread
    • Two-variable data: models and scatterplots
    • Probability and conditional probability
    • Inference from sample statistics and margin of error
    • Evaluating statistical claims: observational studies and Experiments

SAT MAth and English  – full syllabus practice tests

Question Medium

For a science project, Anka recorded whether it rained each weekday and weekend day for 12 weeks. Her results are summarized in the table below.

 RainNo rainTotal
Number of weekdays124860
Number of weekend days81624
Total206484

If one of the days on which there was no rain is selected at random, what is the probability the day was a weekend day?

A) \(\frac{4}{21}\)

B) \(\frac{1}{4}\)

C) \(\frac{2}{3}\)

D) \(\frac{3}{4}\)

▶️ Answer/Explanation
Solution

Answer: B

Days with no rain = 64, weekend days with no rain = 16. Probability = \(\frac{16}{64} = \frac{1}{4}\).

Question Medium

The figure shown is divided into 100 squares of equal area, where 60 squares are shaded.

Shaded square grid

If one of these squares is selected at random how much greater is the probability of selecting a shaded square than the probability of selecting a square that is not shaded?

A) 0.20

B) 0.40

C) 0.60

D) 0.80

▶️ Answer/Explanation
Solution

Answer: A

Probability of shaded square = \(\frac{60}{100} = 0.6\), probability of unshaded square = \(\frac{40}{100} = 0.4\). Difference = \(0.6 – 0.4 = 0.2\).

Question Medium

According to the 2010 Census, the adult population aged 18 or greater of the United States in 2010 was 234,564,071. In 2010, a survey was conducted among a randomly chosen sample of adults aged 18 or greater in the United States about their preference to live in a warm climate or a cool climate. The table below displays a summary of the survey results.

 WarmCoolNo preferenceTotal
18-35 years old29516845508
36-50 years old24612341410
51-65 years old23811748403
Greater than 65 years old1377864279
Total9164861981,600

Which of the following is closest to the difference between the percentage of adults aged 18-50 years who responded “warm” and the percentage of adults aged 51 years or greater who responded “warm”?

  1. 4%
  2. 5%
  3. 10%
  4. 18%
▶️ Answer/Explanation
Solution

Answer: A

Percentage of 18-50 years old preferring “warm” = \(\frac{295 + 246}{508 + 410} \times 100 \approx 54.3\%\).

Percentage of 51+ years old preferring “warm” = \(\frac{238 + 137}{403 + 279} \times 100 \approx 50.6\%\).

Difference ≈ \(54.3\% – 50.6\% = 3.7\%\), closest to 4%.

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