Digital SAT Math Practice Questions - Advanced : Probability and conditional probability - New Syllabus
DSAT 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
▶️Last Minutes DSAT Math revision Sheet
DSAT MAth and English – full syllabus practice tests
▶️ Answer/Explanation
Ans: C
Given: \( P(Y) = 2P(B) \), \( P(Y) = 6P(R) \), \( P(R) + P(B) + P(Y) = 1 \)
From \( P(Y) = 6P(R) \), \( P(R) = \frac{P(Y)}{6} \)
From \( P(Y) = 2P(B) \), substitute: \( P(R) = \frac{2P(B)}{6} = \frac{P(B)}{3} \)
Substitute into sum: \( \frac{P(B)}{3} + P(B) + 2P(B) = 1 \)
Combine: \( \frac{P(B)}{3} + \frac{9P(B)}{3} = 1 \), \( \frac{10P(B)}{3} = 1 \)
Solve: \( 10P(B) = 3 \), \( P(B) = \frac{3}{10} = 0.3 \)
Question Hard
The table summarizes the distribution of age and assigned group for participants in a study.
One of these participants will be selected at random. What is the probability of selecting a participant from group A, given that the participant is at least 10 years of age? (Express your answer as a decimal or fraction, not as a percent.)
▶️Answer/Explanation
Ans: 1 / 3
First, let’s determine the total number of participants who are at least 10 years old. These participants are in the \(10-19\) years and \(20+\) years age groups. From the table, we have:
The total number of participants who are at least 10 years old is:
\[
(18+22)+(14+32)+(28+6)=40+46+34=120
\]
Next, let’s determine the number of participants from Group A who are at least 10 years old:
\[
18(10-19 \text { years })+22(20+\text { years })=40
\]
The probability of selecting a participant from Group A given that the participant is at least 10 years old is the ratio of the number of Group A participants who are at least 10 years old to the total number of participants who are at least 10 years old:
\[
\text { Probability }=\frac{\text { Number of Group A participants at least } 10 \text { years old }}{\text { Total number of participants at least } 10 \text { years old }}=\frac{40}{120}=\frac{1}{3}
\]
▶️ Answer/Explanation
Ans: C
Red maples at Site A = 63, Total red maples = 90
Probability = \( \frac{63}{90} = \frac{7}{10} = 0.7 \)
▶️ Answer/Explanation
Ans: B
Total area = \( 7 \times 5 = 35 \) sq in
Smaller rectangle area = \( 3 \times 5 = 15 \) sq in
Shaded area = \( 35 – 15 = 20 \) sq in
Probability = \( \frac{20}{35} = \frac{4}{7} \approx 0.571 \)
Question Hard
The table shows the results of a poll of 1,000 people. Respondents were asked to agree or disagree with the statement “I rely too much on my phone.” If a respondent who was selected at random disagrees with the statement, which of the following is closest to the probability that the respondent selected is at least 45 years old?
A) 0.37
B) 0.45
C) 0.49
D) 0.55
▶️Answer/Explanation
Ans:D
number of respondents who disagree and are at least 45 years old:
\[
\text{Disagree (45-64)} = 201
\]
\[
\text{Disagree (65 and up)} = 102
\]
total number of respondents who disagree and are at least 45 years old:
\[
201 + 102 = 303
\]
total number of respondents who disagree:
\[
\text{Total Disagree} = 548
\]
probability that a respondent who disagrees is at least 45 years old:
\[
\text{Probability} = \frac{303}{548}
\]
\[
\frac{303}{548} \approx 0.552
\]