Continuous Probability and Geometric Probability Short Notes for GATE DA
Continuous Probability and Geometric Probability short notes for GATE DA: 1 study cards covering concepts, formulas, shortcuts and exam traps, plus solved pra
continuous probability and geometric probability short notes
Continuous and Geometric Probability Toolkit
Core Principles
Continuous Probability:P(a≤X≤b)=∫abf(x)dx. Probability of an exact point is 0.
Uniform Distribution U(a,b):
PDF: f(x)=b−a1
Mean: 2a+b
Variance: 12(b−a)2
Geometric Probability:P=Total MeasureFavorable Measure (Length, Area, or Volume).
Exam Methods
Single Uniform Variable: Find the length of the favorable interval and divide by the total length (b−a).
Two Independent Uniform Variables:
Draw the rectangle [a,b]×[c,d]. Total area = (b−a)(d−c).
Shade the region satisfying the condition (e.g., X≥Y).
Calculate the area of the shaded region using geometry (triangles, rectangles).
Probability = Total AreaShaded Area.
Key Traps
Do not confuse discrete uniform (PMF, P(X=x)>0) with continuous uniform (PDF, P(X=x)=0).
In continuous distributions, P(X<c)=P(X≤c). Boundaries do not matter.
For meeting problems, use the complement: Total Area - Unfavorable Area is often easier than calculating the favorable band directly.
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Question 1
Level 1: Warm-up
Two independent random variables X∼U(0,2) and Y∼U(0,3) are given. What is the area of the joint sample space rectangle in the XY-plane?
Question 2
Level 1: Warm-up
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)?
Question 3
Level 1: Warm-up
A random variable X follows a continuous uniform distribution on the interval [0,10]. Which of the following statements is true?
Question 4
Level 1: Warm-up
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?
Question 5
Level 1: Warm-up
A continuous random variable X has probability density function f(x)=kx for 0≤x≤2, and f(x)=0 otherwise. Find the value of k.
Question 6
Level 1: Warm-up
A random variable X follows a continuous uniform distribution on the interval [2,8]. What is the variance of X?
Question 7
Level 1: Warm-up
A point is chosen uniformly at random from the interval [−5,5]. What is the probability that the point satisfies the inequality x2<9?
Question 8
Level 1: Warm-up
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?
Question 9
Level 1: Warm-up
For a continuous random variable X with probability density function f(x), which of the following conditions must always be satisfied?
Question 10
Level 1: Warm-up
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∈[0,10]?
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Continuous Probability and Geometric Probability Short Notes for GATE DA
Continuous Probability and Geometric Probability short notes for GATE DA: 1 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.
Continuous and Geometric Probability Toolkit
Core Principles
Continuous Probability:P(a≤X≤b)=∫abf(x)dx. Probability of an exact point is 0.
Uniform Distribution U(a,b):
PDF: f(x)=b−a1
Mean: 2a+b
Variance: 12(b−a)2
Geometric Probability:P=Total MeasureFavorable Measure (Length, Area, or Volume).
Exam Methods
Single Uniform Variable: Find the length of the favorable interval and divide by the total length (b−a).
Two Independent Uniform Variables:
Draw the rectangle [a,b]×[c,d]. Total area = (b−a)(d−c).
Shade the region satisfying the condition (e.g., X≥Y).
Calculate the area of the shaded region using geometry (triangles, rectangles).
Probability = Total AreaShaded Area.
Key Traps
Do not confuse discrete uniform (PMF, P(X=x)>0) with continuous uniform (PDF, P(X=x)=0).
In continuous distributions, P(X<c)=P(X≤c). Boundaries do not matter.
For meeting problems, use the complement: Total Area - Unfavorable Area is often easier than calculating the favorable band directly.
Continuous Probability and Geometric Probability: Solved Questions with Step-by-Step Explanations (10 Problems)
Question 1 · Probability and StatisticsMCQ
Two independent random variables X∼U(0,2) and Y∼U(0,3) are given. What is the area of the joint sample space rectangle in the XY-plane?
A.
5
B.
6
C.
2.5
D.
1.5
Correct Answer:
B
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∈[0,2], so the width is 2−0=2
Y∈[0,3], so the height is 3−0=3
Step 2: The joint sample space is the rectangle [0,2]×[0,3].
Step 3: Calculate the area:
Area=width×height=2×3=6
Answer: B
Question 2 · Probability and StatisticsMCQ
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:
A
Step-by-Step Solution
Key idea: For a uniform distribution U(a,b), the PDF is constant and equals b−a1.
Step 1: Identify the parameters: a=0, b=10.
Step 2: The PDF of U(a,b) is:
f(x)=b−a1 for a≤x≤b
Step 3: Calculate:
f(x)=10−01=101=0.1
Step 4: Since the PDF is constant over [0,10], the maximum value is 0.1.
Answer: A
Question 3 · Probability and StatisticsMCQ
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:
A
Step-by-Step Solution
Key idea: For a uniform distribution, probability is proportional to interval length. Use the area rule: P(a<X<b)=total lengthb−a.
Step 1: The total interval is [0,10], so total length = 10.
Step 2: Check each option:
Option A:P(X<5)=P(0≤X<5)
- Favorable length = 5−0=5
- P(X<5)=105=0.5 ✓
Option B:P(X>12)
- Since X∈[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)=106=0.6, not 0.8 ✗
Answer: A
Question 4 · Probability and StatisticsMCQ
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:
A
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 = πR2=π(5)2=25π.
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 = πr2=π(2)2=4π.
Step 3: Calculate the probability.
P=Total areaFavorable area=25π4π=254=0.16.
Answer: 0.16
Question 5 · Probability and StatisticsNAT
A continuous random variable X has probability density function f(x)=kx for 0≤x≤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:
∫−∞∞f(x)dx=1
Step 2: Since f(x)=0 outside [0,2]:
∫02kxdx=1
Step 3: Evaluate the integral:
k∫02xdx=k[2x2]02=k(24−0)=2k
Step 4: Solve for k:
2k=1⟹k=0.5
Answer: 0.50
Question 6 · Probability and StatisticsMCQ
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:
B
Step-by-Step Solution
Key idea: For a uniform distribution U(a,b), the variance is 12(b−a)2.
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)=12(b−a)2=1262=1236=3
Answer: B
Question 7 · Probability and StatisticsMCQ
A point is chosen uniformly at random from the interval [−5,5]. What is the probability that the point satisfies the inequality x2<9?
A.
0.30
B.
0.60
C.
0.80
D.
0.90
Correct Answer:
B
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.
x2<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=Total lengthFavorable length=106=0.60.
Answer: 0.60
Question 8 · Probability and StatisticsMCQ
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:
B
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 StatisticsMCQ
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.
∫−∞∞f(x)dx=1
C.
P(X=x)=f(x) for all x
D.
f(x)≤1 for all x
Correct Answer:
B
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)≥0 for all x (non-negative, not strictly positive)
∫−∞∞f(x)dx=1 (total area equals 1)
Step 2: Check each option:
Option A: Wrong because f(x)≥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 StatisticsMCQ
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∈[0,10]?
A.
0.0
B.
0.1
C.
0.5
D.
1.0
Correct Answer:
A
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 x0, P(X=x0)=0.
Step 3: Determine the maximum value.
Since P(X=x)=0 for every x∈[0,10], the maximum value of this probability over the interval is simply 0.