Continuous Probability and Geometric Probability Practice Questions for GATE DA

Solve 21+ Continuous Probability and Geometric Probability practice questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Continuous Probability and Geometric Probability

Your journey through this chapter

Continuous Random Variables and PDFs

From counting discrete outcomes to measuring continuous ones.

The Continuous Uniform Distribution

The simplest continuous model, where every interval of the same length is equally likely.

Geometric Probability

When chance meets geometry: lengths, areas, and volumes.

Joint Distributions and Area Methods

Solving multi-variable problems by calculating areas in the plane.

End goal: visualize probability as area under a curve or as a ratio of geometric measures, and solve continuous probability questions effortlessly.

The Shift to Continuous: From Counting to Measuring

The fundamental shift

Discrete: Outcomes are countable. We use a Probability Mass Function (PMF). $P (X = x) > 0$ .
Continuous: Outcomes are measurements (time, length, area). We use a Probability Density Function (PDF). $P (X = x) = 0$ .

The Probability Density Function (PDF)

For a continuous random variable $X$ , the PDF $f (x)$ satisfies:

$f (x) \geq 0$ for all $x$ .
The total area under the curve is 1: $\int_{- \infty}^{\infty} f (x) d x = 1$ .

How to find probabilities

The probability that $X$ falls in an interval $[a, b]$ is the area under the PDF over that interval:

P (a \leq X \leq b) = \int_{a}^{b} f (x) d x

Key intuition: In continuous probability, you do not calculate the probability of a point; you calculate the probability of a region by measuring its area.

Continuous Probability and Geometric Probability: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Probability and Statistics MCQ

Two independent random variables $X \sim U (0, 2)$ and $Y \sim U (0, 3)$ are given. What is the area of the joint sample space rectangle in the $X Y$ -plane?

A.
5
B.
6
C.
2.5
D.
1.5

Correct Answer:

Step-by-Step Solution

Key idea: For independent uniform random variables, the joint sample space is a rectangle with dimensions equal to the interval lengths.

Step 1: Identify the intervals:

$X \in [0, 2]$ , so the width is $2 - 0 = 2$
$Y \in [0, 3]$ , so the height is $3 - 0 = 3$

Step 2: The joint sample space is the rectangle $[0, 2] \times [0, 3]$ .

Step 3: Calculate the area:

$Area = width \times height = 2 \times 3 = 6$

Answer: B

Question 2 · Probability and Statistics MCQ

A random variable $X$ follows a continuous uniform distribution on the interval $[0, 10]$ . What is the maximum value of its probability density function $f (x)$ ?

A.
0.1
B.
0.5
C.
1
D.
10

Correct Answer:

Step-by-Step Solution

Key idea: For a uniform distribution $U (a, b)$ , the PDF is constant and equals $\frac{1}{b - a}$ .

Step 1: Identify the parameters: $a = 0$ , $b = 10$ .

Step 2: The PDF of $U (a, b)$ is:

$f (x) = \frac{1}{b - a} for a \leq x \leq b$

Step 3: Calculate:

$f (x) = \frac{1}{10 - 0} = \frac{1}{10} = 0.1$

Step 4: Since the PDF is constant over $[0, 10]$ , the maximum value is 0.1.

Answer: A

Question 3 · Probability and Statistics MCQ

A random variable $X$ follows a continuous uniform distribution on the interval $[0, 10]$ . Which of the following statements is true?

A.
$P (X < 5) = 0.5$
B.
$P (X > 12) = 0.2$
C.
$P (X = 5) = 0.1$
D.
$P (2 < X < 8) = 0.8$

Correct Answer:

Step-by-Step Solution

Key idea: For a uniform distribution, probability is proportional to interval length. Use the area rule:

P (a < X < b) = \frac{b - a}{total length}

. Step 1: The total interval is

[0, 10]

, so total length = 10. Step 2: Check each option: Option A:

P (X < 5) = P (0 \leq X < 5)

- Favorable length =

5 - 0 = 5

P (X < 5) = \frac{5}{10} = 0.5

✓ Option B:

P (X > 12)

- Since

X \in [0, 10]

X

cannot exceed 10 -

P (X > 12) = 0

, not 0.2 ✗ Option C:

P (X = 5)

- For continuous distributions,

P (X = exact value) = 0

P (X = 5) = 0

, not 0.1 ✗ Option D:

P (2 < X < 8)

- Favorable length =

8 - 2 = 6

P (2 < X < 8) = \frac{6}{10} = 0.6

, not 0.8 ✗ Answer: A

Question 4 · Probability and Statistics MCQ

A dart is thrown at a circular board of radius 5. Assuming the dart lands uniformly at random on the board, what is the probability that it lands within a distance of 2 from the center?

A.
0.16
B.
0.40
C.
0.08
D.
0.20

Correct Answer:

Step-by-Step Solution

Key idea: This is a 2D geometric probability question. The probability is the ratio of the favorable area to the total area.

Step 1: Identify the total region. The board is a circle of radius $R = 5$ .

Total area = $π R^{2} = π (5)^{2} = 25 π$ .

Step 2: Identify the favorable region. The dart must land within a distance of $r = 2$ from the center. This is a smaller circle of radius 2.

Favorable area = $π r^{2} = π (2)^{2} = 4 π$ .

Step 3: Calculate the probability.

$P = \frac{Favorable area}{Total area} = \frac{4 π}{25 π} = \frac{4}{25} = 0.16$ .

Answer: 0.16

Question 5 · Probability and Statistics NAT

A continuous random variable $X$ has probability density function $f (x) = k x$ for $0 \leq x \leq 2$ , and $f (x) = 0$ otherwise. Find the value of $k$ .

Correct Answer:

0.50

Step-by-Step Solution

Key idea: This is a normalization problem. Use the property that the total area under a PDF equals 1.

Step 1: Set up the normalization equation:

$\int_{- \infty}^{\infty} f (x) d x = 1$

Step 2: Since $f (x) = 0$ outside $[0, 2]$ :

$\int_{0}^{2} k x d x = 1$

Step 3: Evaluate the integral:

$k \int_{0}^{2} x d x = k [\frac{x ^{2}}{2}]_{0}^{2} = k (\frac{4}{2} - 0) = 2 k$

Step 4: Solve for $k$ :

$2 k = 1 ⟹ k = 0.5$

Answer: 0.50

Question 6 · Probability and Statistics MCQ

A random variable $X$ follows a continuous uniform distribution on the interval $[2, 8]$ . What is the variance of $X$ ?

A.
2
B.
3
C.
4
D.
6

Correct Answer:

Step-by-Step Solution

Key idea: For a uniform distribution $U (a, b)$ , the variance is $\frac{( b - a ) ^{2}}{12}$ .

Step 1: Identify the parameters: $a = 2$ , $b = 8$ .

Step 2: Calculate the interval length:

$b - a = 8 - 2 = 6$

Step 3: Apply the variance formula:

$Var (X) = \frac{( b - a ) ^{2}}{12} = \frac{6 ^{2}}{12} = \frac{36}{12} = 3$

Answer: B

Question 7 · Probability and Statistics MCQ

A point is chosen uniformly at random from the interval $[- 5, 5]$ . What is the probability that the point satisfies the inequality $x^{2} < 9$ ?

A.
0.30
B.
0.60
C.
0.80
D.
0.90

Correct Answer:

Step-by-Step Solution

Key idea: This is a 1D geometric probability problem. The probability is the ratio of the length of the favorable interval to the total interval length.

Step 1: Identify the total interval. The point is chosen from $[- 5, 5]$ .

Total length = $5 - (- 5) = 10$ .

Step 2: Solve the inequality to find the favorable region.

$x^{2} < 9 ⟹ - 3 < x < 3$ .

The favorable interval is $(- 3, 3)$ .

Step 3: Calculate the length of the favorable interval.

Favorable length = $3 - (- 3) = 6$ .

Step 4: Calculate the probability.

$P = \frac{Favorable length}{Total length} = \frac{6}{10} = 0.60$ .

Answer: 0.60

Question 8 · Probability and Statistics MCQ

Consider the following statements regarding geometric probability:

Assertion (A): For a point chosen at random on a line segment of length $L$ , the probability of it falling in a sub-segment of length $l$ is $l / L$ .

Reason (R): The geometric measure of a 1D region is its area, so the probability is calculated as the ratio of areas.

Which of the following is correct?

A.
Both A and R are true, and R is the correct explanation of A.
B.
A is true, but R is false.
C.
A is false, but R is true.
D.
Both A and R are false.

Correct Answer:

Step-by-Step Solution

Key idea: This question tests the understanding of geometric measures in different dimensions. We must evaluate both the assertion and the reason independently.

Step 1: Evaluate Assertion (A).

The sample space is a line segment, which is a 1D region. The probability is indeed the ratio of the favorable length to the total length: $P = l / L$ .

So, Assertion (A) is True.

Step 2: Evaluate Reason (R).

The reason states that the geometric measure of a 1D region is its "area".

This is incorrect. The measure of a 1D region is its "length". Area is the measure for 2D regions.

So, Reason (R) is False.

Step 3: Combine the evaluations.

A is true, but R is false.

Answer: A is true, but R is false.

Question 9 · Probability and Statistics MCQ

For a continuous random variable $X$ with probability density function $f (x)$ , which of the following conditions must always be satisfied?

A.
$f (x) > 0$ for all $x$
B.
$\int_{- \infty}^{\infty} f (x) d x = 1$
C.
$P (X = x) = f (x)$ for all $x$
D.
$f (x) \leq 1$ for all $x$

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct recall question about the fundamental properties of a PDF.

Step 1: Recall the two defining properties of a PDF $f (x)$ :

$f (x) \geq 0$ for all $x$ (non-negative, not strictly positive)
$\int_{- \infty}^{\infty} f (x) d x = 1$ (total area equals 1)

Step 2: Check each option:

Option A: Wrong because $f (x) \geq 0$ , not $f (x) > 0$ . A PDF can be zero in some regions.
Option B: Correct. This is the normalization condition.
Option C: Wrong. For continuous random variables, $P (X = x) = 0$ , not $f (x)$ .
Option D: Wrong. A PDF can exceed 1. For example, $U (0, 0.5)$ has $f (x) = 2$ .

Answer: B

Question 10 · Probability and Statistics MCQ

Let $X$ be a continuous random variable uniformly distributed on the interval $[0, 10]$ . What is the maximum value of the probability $P (X = x)$ over all $x \in [0, 10]$ ?

A.
0.0
B.
0.1
C.
0.5
D.
1.0

Correct Answer:

Step-by-Step Solution

Key idea: This question tests the fundamental difference between discrete and continuous random variables. For any continuous random variable, the probability of it taking any exact single value is zero.

Step 1: Identify the type of random variable. $X$ is a continuous random variable.

Step 2: Recall the property of continuous random variables.

For any continuous random variable $X$ and any specific real number $x_{0}$ , $P (X = x_{0}) = 0$ .

Step 3: Determine the maximum value.

Since $P (X = x) = 0$ for every $x \in [0, 10]$ , the maximum value of this probability over the interval is simply 0.

Answer: 0.0

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Continuous Probability and Geometric Probability Practice Questions for GATE DA

Chapter Roadmap: Continuous Probability and Geometric Probability

The Shift to Continuous: From Counting to Measuring

The fundamental shift

The Probability Density Function (PDF)

How to find probabilities

Continuous Probability and Geometric Probability: Solved Questions with Step-by-Step Explanations (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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