A random variable attaches a number to each outcome. Example: defective chips in a batch. You compute probabilities with three tools. The PMF lists each value's probability for a discrete variable. The PDF gives probability as area under a curve for a continuous variable. The CDF gives the probability that X is at most x.
A problem belongs here when it gives a probability table, a density, or an F(x) and asks for the probability of a range.
You will practice: PMF sums, PDF integrals, CDF conversions, max and min.
12 problems, each with a worked answer. Verified by an independent second solve.
Work them on paper, in order; difficulty ramps gently.
1.
Basic
A quality-control station tests microchips in batches of 3. The number of defective chips X in a batch has PMF p(0) = 0.5, p(1) = 0.3, p(2) = c, p(3) = 0.05 for some constant c. Find P(X ≥ 2).
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Answer: 0.2
The PMF must sum to 1, which fixes c. Then add the two top masses.
c=1−(0.5+0.3+0.05)=0.15
P(X≥2)=p(2)+p(3)=0.15+0.05=0.20
2.
Basic
The lifetime T (in years) of a temperature sensor has PDF f(t) = t/8 for 0 ≤ t ≤ 4 and f(t) = 0 otherwise. Find the probability that a sensor fails within its first 2 years, P(T ≤ 2).
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Answer: 41
Integrate the PDF from 0 to 2.
P(T≤2)=∫028tdt=[16t2]02=164=41
3.
Basic
In a football league, the number of goals X that Rovers score in a match has PMF p(0) = 0.2, p(1) = 0.35, p(2) = 0.25, p(3) = 0.15, p(4) = 0.05. Find the probability that Rovers score at least 2 goals in a match.
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Answer: 0.45
At least 2 goals means X equals 2, 3, or 4, so sum those point masses. As a check, all five masses sum to 1, so the PMF is valid.
P(X≥2)=0.25+0.15+0.05=0.45
4.
Basic
A delivery drone's flight time X (in minutes) has PDF f(x) = 2x/25 for 0 ≤ x ≤ 5 and f(x) = 0 otherwise. Find the probability that a flight lasts more than 3 minutes.
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Answer: 2516
Probability is the area under the density beyond 3. Equivalently take 1 minus the area up to 3.
P(X>3)=∫35252xdx=2525−9=2516
5.
Basic
At a ceramics studio, the number of cracked mugs X in one kiln firing has PMF p(0) = 0.4, p(1) = k, p(2) = 0.15, p(3) = 0.05. Find P(X ≤ 1).
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Answer: 0.8
The PMF must sum to 1, which fixes k. Then add the first two masses.
k=1−(0.4+0.15+0.05)=0.4
P(X≤1)=p(0)+p(1)=0.4+0.4=0.8
6.
Basic
The wait time X (in minutes) in line at a food truck has PDF f(x) = 1/6 for 0 ≤ x ≤ 6 and f(x) = 0 otherwise. Find P(2 ≤ X ≤ 5).
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Answer: 21
Probability is the area under the flat PDF over the interval from 2 to 5.
P(2≤X≤5)=(5−2)⋅61=63=21
7.
Basic
A campus bike-share station records the number of helmets X borrowed with each rental. X has PMF p(0) = 0.6, p(1) = 0.3, and p(2) is the only other possible value. Find p(2).
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Answer: 0.1
The point masses must sum to 1.
p(2)=1−0.6−0.3=0.1
8.
Basic
The time X (in minutes) you wait for an elevator has CDF F(x) = x/5 for 0 ≤ x ≤ 5, with F(x) = 0 for x < 0 and F(x) = 1 for x > 5. Find P(X ≤ 2).
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Answer: 52
The CDF gives P(X ≤ x) directly, so plug in x = 2.
F(2)=52=0.4
9.
Basic
The daily amount of water X (in liters) a potted fern absorbs has PDF f(x) = x/2 for 0 ≤ x ≤ 2 and f(x) = 0 otherwise. Find P(X ≤ 1).
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Answer: 41
Probability is the area under the PDF from 0 to 1.
∫012xdx=[4x2]01=41
10.
Basic
At a cafe, the number of espresso shots X in a random customer's drink has PMF p(1) = 0.5, p(2) = 0.3, p(3) = 0.15, p(4) = 0.05. Find P(X ≥ 3).
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Answer: 0.2
Add the point masses at 3 and 4.
P(X≥3)=0.15+0.05=0.2
11.
Basic
The time X (in minutes) a solver spends on one crossword clue has CDF F(x) = x³/27 for 0 ≤ x ≤ 3, with F(x) = 0 for x < 0 and F(x) = 1 for x > 3. Find P(1 ≤ X ≤ 2).
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Answer: 277
An interval probability is a difference of CDF values.
F(2)−F(1)=278−271=277
12.
Basic
At an orchard stand, the number X of apple varieties in a sampler box takes the values 1, 2, 3, 4 with P(X = x) = cx for some constant c. Find P(X = 3).
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Answer: 103
Normalize first: the masses c, 2c, 3c, 4c must sum to 1.
c(1+2+3+4)=10c=1⇒c=101
P(X=3)=3c=103