01 · One or fewer
For the same X, find F(1).
Hint
Add the masses at 0 and 1.
Worked solution
0.32768+0.4096=0.73728.
Understand · explore · practise
Cumulative binomial probabilities
Build cumulative binomial probabilities by adding point masses, read a cumulative table and choose the correct inclusive cutoff on a calculator.
Before you startBinomial point probabilities and integer-valued counts.
01 / Build a cumulative total
A cumulative probability includes every count up to its cutoff.
F(k)=P(X≤k).
For integer k within the support, add P(X=0) through P(X=k), including the endpoint.
For a binomial count, there are no negative values. Cumulative does not mean the probability of exactly k: it combines all qualifying counts.
Add the bars through the cutoffExplore
X~B(5,0.2). Choose an integer cutoff k. The highlighted masses contribute to F(k)=P(X≤k).
At k=2, add the masses for 0, 1 and 2: F(2)=0.94208.
02 / Add disjoint count events
Exactly zero, one and two successes cannot happen together.
Watch: three masses make one cumulative total
Pause, replay or seek freely. The notes explain the same idea and stay in view.
X~B(5,0.2). Find P(X≤2).Worked example
P(X=0)=0.32768; P(X=1)=0.4096
These are separate point probabilities.
P(X=2)=0.2048
There are C(5,2)=10 possible success-position pairs.
F(2) = 0.32768 + 0.4096 + 0.2048 = 0.94208
Include the mass at 2.
02 · Exactly two
Does F(2)=0.94208 answer P(X=2)?
Hint
Which counts did F(2) include?
Worked solution
No. P(X=2)=0.2048; F(2) includes 0, 1 and 2.
03 / Read the distribution and the endpoint
A cumulative table has to match n and p.
Suppose a table is explicitly labelled F(k)=P(X≤k) for Y~B(6,0.4). Its values are shown below. Read the stated n and p before choosing the k row; a neighbouring probability column describes a different model.
| k | F(k) |
|---|---|
| 0 | 0.046656 |
| 1 | 0.233280 |
| 2 | 0.544320 |
| 3 | 0.820800 |
| 4 | 0.959040 |
| 5 | 0.995904 |
| 6 | 1.000000 |
03 · Read a total
For Y~B(6,0.4), find P(Y≤3).
Hint
Use the k=3 entry.
Worked solution
0.820800.
04 · Read in words
What event has probability 0.959040 in that table?
Hint
Which endpoint belongs to that entry?
Worked solution
Y≤4: at most four successes in six trials.
05 · Wrong parameter
Can the same values be used for B(6,0.5)?
Hint
The probabilities depend on p as well as n.
Worked solution
No. Recalculate or use the table for p=0.5.
04 / Specify the cumulative calculation
Check what your calculator means by its output.
Choose the binomial cumulative function and enter the labels n, p and upper cutoff k. Some interfaces request lower and upper limits: for P(X≤k), use lower limit 0. Read the calculator’s displayed event; do not rely on the order another model uses for its inputs.
For Y~B(6,0.4), calculate F(2).Worked example
Use n=6, p=0.4, upper count 2
This asks for P(Y≤2).
The cumulative result is 0.54432
A point result of 0.31104 would answer P(Y=2) instead.
06 · Diagnose a display
An answer of 0.31104 is obtained for F(2). Explain the likely mistake.
Hint
Compare the point mass at 2.
Worked solution
The point-probability function may have been selected. Add all counts 0,1,2 or use the cumulative function.
07 · Round once
Give F(2)=0.54432 to four decimal places.
Hint
Inspect the fifth decimal digit.
Worked solution
0.5443. Retain the full value if it is used in another calculation.
05 / Use the support before calculating
A cumulative total becomes zero or one outside the possible counts.
For X~B(n,p), F(k)=0 when k<0, and F(k)=1 when k≥n.
Every possible count is an integer from 0 through n; some masses can be zero when p=0 or p=1.
08 · Below the support
For X~B(5,0.2), find P(X≤−1).
Hint
There are no negative counts.
Worked solution
0.
09 · Include every count
For the same X, find F(5) and F(6).
Hint
Both cutoffs include all possible values.
Worked solution
Both are 1.
10 · Zero is possible
For that X, is F(0)=0?
Hint
F(0) includes the mass at zero.
Worked solution
No. F(0)=P(X=0)=0.8⁵=0.32768.
06 / A count cannot sit between integers
Interpret a noninteger cutoff using the included counts.
For integer-valued X, P(X≤2.7)=P(X≤2), because no possible count lies strictly between 2 and 2.7. State the event first; if a calculator requires an integer cutoff, enter the largest included integer.
11 · Noninteger boundary
For X~B(5,0.2), find P(X≤1.8).
Hint
The included counts are 0 and 1.
Worked solution
F(1)=0.73728.
12 · Exactly at the boundary
Are P(X≤2) and P(X<2) equal?
Hint
Does the mass at 2 appear in both?
Worked solution
No. P(X≤2)=0.94208 and P(X<2)=F(1)=0.73728. Their difference is P(X=2)=0.2048.
07 / Check the direction and size
Cumulative totals never decrease as the cutoff increases.
Adding another nonnegative mass cannot reduce the total. A cumulative table starts at zero below the support and reaches one at the top. Adjacent entries can be equal when the intervening point mass is zero.
13 · Impossible table
A proposed cumulative table gives F(2)=0.7 and F(3)=0.6. Can it be correct?
Hint
Increasing the cutoff adds possible outcomes.
Worked solution
No. A cumulative probability cannot decrease.
14 · Recover a mass
For Y~B(6,0.4), use F(3)=0.8208 and F(2)=0.54432 to find P(Y=3).
Hint
Subtract the total through 2 from the total through 3.
Worked solution
0.8208−0.54432=0.27648.
08 / Model, endpoint, inclusive sum
Read the event before pressing a calculator button.
Identify n and p, list which integer counts qualify and use F(k)=P(X≤k). Check the table or calculator is cumulative and includes the endpoint. Keep unrounded values for later subtraction or complements.
Section 1 of 8 · Build a cumulative total