Interval Probabilities Under a Uniform Model

Why Interval Probabilities Are Length Ratios

For a continuous uniform random variable, probability is spread evenly across its support. This makes interval probabilities a geometric calculation: find how much of the interval lies within the support, then divide that length by the support’s total length. The main care required is to identify the overlap correctly, especially when an event extends beyond the support or is described using different units.

Definition: A continuous random variable $X$ has a uniform distribution on $[a,b]$, written $X\sim\operatorname{Uniform}(a,b)$, when $a<b$ and its density is $f_X(x)=1/(b-a)$ for $a\le x\le b$, and $f_X(x)=0$ otherwise. The support is $[a,b]$, with total length $b-a$.

Because the density is constant on the support, every subinterval of a given length has the same probability. An interval of length $L$ contained in $[a,b]$ therefore has probability $L/(b-a)$. If an event interval partly lies outside the support, only its overlap with $[a,b]$ contributes probability.

Theorem: For $X\sim\operatorname{Uniform}(a,b)$, the probability that $X$ belongs to an interval is the length of that interval’s intersection with $[a,b]$, divided by $b-a$. In particular, for $c<d$, $$ P(c<X<d)= \frac{\max\!\left(0,\ \min(d,b)-\max(c,a)\right)}{b-a}. $$ The same probability is obtained if either or both inequality signs are changed to include an endpoint.

The maximum and minimum in the formula perform the clipping: $\max(c,a)$ is the left edge of the overlap, and $\min(d,b)$ is its right edge. If the right edge is not greater than the left edge, there is no overlap of positive length and the probability is zero. The formula also handles intervals that fully contain the support, giving probability one.

Justification of the Length Formula

For any interval lying within the support, the probability is the integral of the density over that interval. Since the density is constant, the integral is the density height multiplied by the interval’s length. For an interval that reaches beyond the support, the density is zero outside $[a,b]$, so only the overlapping part contributes.

P(c<X<d) =\int_{\max(c,a)}^{\min(d,b)}\frac{1}{b-a}\,dx =\frac{\min(d,b)-\max(c,a)}{b-a}

This integral applies when the overlap has positive length. When there is no positive-length overlap, the probability is zero, which accounts for the maximum with zero in the general formula. The formula depends on the uniform density being constant on $[a,b]$; for a nonuniform model, interval length alone is not enough, and the density must be integrated over the relevant interval.

A Reliable Calculation Procedure

A useful way to prevent errors is to treat the support and the event as two intervals, find their overlap, and only then calculate a ratio. Do not use the length of the event interval before checking whether all of it is possible under the model.

State the support.
Record the model’s lower and upper endpoints, $a$ and $b$, and calculate its length $b-a$.

Translate the event.
Write the requested interval or intervals clearly. Convert units first if the event and model use different units.

Find the overlap.
Clip each event interval to the support. An interval entirely outside the support contributes zero length.

Divide by the support length.
For one interval, divide its overlap length by $b-a$. For disjoint event intervals, add their overlap lengths before dividing.

Check the result.
A probability must be between zero and one. The total contributing length cannot exceed $b-a$.

Worked Examples

Worked Example: An Interval Extending Below the Support

A hypothetical equipment check takes a time $X$ uniformly distributed from 10 to 34 minutes. Find the probability that the check takes more than 4 minutes but no more than 22 minutes.

Method. Use the overlap-length formula. The model is $X\sim\operatorname{Uniform}(10,34)$, so the support length is $34-10=24$ minutes. The requested event is $4<X\le22$. Its overlap with the support is $10<X\le22$, of length $22-10=12$ minutes.

P(4<X\le22) =\frac{22-10}{34-10} =\frac{12}{24} =0.5000

The probability is $0.5000$. The lower event boundary at 4 minutes is outside the support, so it does not increase the overlap length. The endpoint at 10 minutes has probability zero, as does any single endpoint in this continuous model.

Common mistake avoided: Dividing the event’s untrimmed length, $22-4=18$, by 24 would give $0.7500$. That incorrectly assigns probability to times from 4 to 10 minutes, which are impossible under the stated model. The overlap calculation gives a valid result between zero and one and uses only values in the support.

Worked Example: A Union of Intervals with One Beyond the Support

For a hypothetical covered loss amount $X$, assume $X\sim\operatorname{Uniform}(0,8000)$ dollars. Find the probability that the loss is either between \$1,500 and \$3,000, or at least \$6,000 but no more than \$10,000.

Method. The support length is $8000-0=8000$ dollars. The first event interval, from \$1,500 to \$3,000, lies wholly within the support and has length $3000-1500=1500$ dollars.

The second interval is from \$6,000 to \$10,000, but the model has no losses above \$8,000. Its overlap with the support runs from \$6,000 to \$8,000 and has length $8000-6000=2000$ dollars. The two contributing intervals are disjoint, so their lengths can be added.

P\bigl((1500<X<3000)\ \text{or}\ (6000\le X\le10000)\bigr) =\frac{1500+2000}{8000} =\frac{3500}{8000} =0.4375

The probability is $0.4375$. Including or excluding the listed endpoints does not change the result, because a single point has probability zero.

Common mistake avoided: Using the second interval’s full length of $10000-6000=4000$ dollars would give $(1500+4000)/8000=0.6875$. The extra \$2,000 lies outside the support and cannot contribute probability. The corrected total overlap length is \$3,500, less than the \$8,000 support length, as required.

Worked Example: Convert Units Before Finding the Overlap

A hypothetical processing time $X$ is uniformly distributed from 1.5 to 5.5 hours. Find the probability that the time is between 150 and 270 minutes, inclusive, or exceeds 4.75 hours.

Method. Use a single unit throughout. Convert the interval in minutes to hours: $150/60=2.5$ hours and $270/60=4.5$ hours. The first event interval is therefore from 2.5 to 4.5 hours, with length $4.5-2.5=2$ hours. It lies within the support.

The second event is $X>4.75$ hours. Its overlap with the support runs from 4.75 to 5.5 hours and has length $5.5-4.75=0.75$ hours. These event intervals do not overlap, so the total event length is $2+0.75=2.75$ hours. The support length is $5.5-1.5=4$ hours.

P\bigl((150\le60X\le270)\ \text{or}\ (X>4.75)\bigr) =\frac{2.75}{4} =0.6875

The probability is $0.6875$. As a check, the total contributing length of 2.75 hours is less than the 4-hour support length.

Common mistake avoided: Comparing 150 and 270 directly with boundaries measured in hours mixes units and does not define a meaningful overlap. Convert the event boundaries to hours first; then all lengths in the ratio use the same unit.

Disjoint Intervals, Endpoints, and Checks

For an event consisting of several disjoint intervals, find each interval’s overlap with the support and add the overlap lengths. If the event intervals overlap one another, first express their union without counting any part twice. A total length greater than $b-a$ signals either overlap being double-counted or an incorrect endpoint calculation.

For a continuous uniform variable, the probability of any single value is zero. Consequently, $P(c<X<d)$, $P(c\le X<d)$, $P(c<X\le d)$, and $P(c\le X\le d)$ are equal whenever the endpoints are the same. This does not mean endpoints can be ignored when finding the support or overlap: the interval boundaries still determine the length. It means only that including or excluding those individual points does not change the probability.

Key takeaway: A uniform-model interval probability is an overlap length divided by the support length. Clip event intervals to the support, use consistent units, and count only the length of the event’s union.

Check Your Understanding

For each problem, identify the overlap with the stated support before calculating the probability. Answers are rounded to four decimal places when needed.

A sensor’s reading $X$ is uniformly distributed on $[-4,8]$. Find $P(-7<X\le2)$, explaining briefly why the part below $-4$ does not contribute. (Answer: $0.5000$)
A hypothetical service duration $X$ is uniform on $[20,50]$ minutes. Find the probability that $X<29$ or $42<X\le60$. (Answer: $0.5667$)
A waiting time $X$ is uniform on $[0,12]$ minutes. Find the probability that $X$ is at least 3 minutes and at most 9 minutes, and also exceeds 5 minutes. (Answer: $0.3333$)
A hypothetical repair time $X$ is uniform on $[2,10]$ hours. Find the probability that the time is between 210 and 510 minutes, inclusive. Convert the event boundaries before calculating. (Answer: $0.6250$)
A hypothetical loss amount $X$ is uniform on $[100,700]$ dollars. Find the probability that the loss is at most \$160 or at least \$640. (Answer: $0.2000$)

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