01 · Read the formula
For this PMF, find P(X=2) and P(X=3).
Hint
Use the formula only on its stated support.
Worked solution
P(X=2)=3/9=1/3. P(X=3)=0 because 3 is outside the support.
Understand · explore · practise
Probability mass functions
Check probability tables, find normalising constants, handle piecewise formulas and calculate events from a discrete probability mass function.
Before you startDiscrete random variables, fractions and simple algebra.
01 / Check both requirements
A total of one is necessary, but not sufficient.
For every supported value x, P(X=x) ≥ 0. The sum over all supported values is 1.
These two conditions together also ensure every individual mass is at most 1.
A probability mass function, or PMF, assigns a probability to each possible numerical value. Always state the values for which the formula applies; outside them the mass is zero.
Is this a probability distribution?Explore
Each mass must be nonnegative and all masses together must sum to 1.
02 / Turn weights into probabilities
Divide each nonnegative weight by the total weight.
Watch: normalise unequal weights
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Let P(X=x)=k(x+1) for x in {0,2,4}, and zero otherwise.Worked example
The weights are 1,3,5, with total 9
The probability total is k+3k+5k=9k.
9k=1, so k=1/9
Normalising uses every supported value once.
Masses: 1/9,1/3,5/9
They are nonnegative and sum to 1.
02 · An upper event
Find P(X≥2).
Hint
Include 2 and 4.
Worked solution
1/3+5/9=8/9.
03 / Missing values in a table
Account for every occurrence of a parameter.
03 · One missing mass
A table on x=0,1,2,3 has masses 0.15,0.25,a,0.30. Find a.
Hint
Subtract the known total from 1.
Worked solution
a=1−0.70=0.30. The resulting masses are all nonnegative.
04 · A repeated mass
A table on x=−1,0,2,5 has masses b,0.2,b,0.3. Find b.
Hint
There are two entries equal to b.
Worked solution
2b+0.5=1, giving b=0.25.
05 · Formula as weights
P(Y=y)=cy² on y=1,2,3. Find c and P(Y>1).
Hint
The weights are 1,4,9.
Worked solution
14c=1, so c=1/14. P(Y>1)=(4+9)/14=13/14.
04 / Use the correct branch
A piecewise PMF still describes one distribution.
P(Z=z)=kz for z=1 or4; P(Z=2)=2k; zero elsewhere.Worked example
Write the table before summing
At1 the mass is k, at 2 it is 2k, at 4 it is 4k.
k+2k+4k=1
So k=1/7.
P(Z<4)=1/7+2/7=3/7
The condition selects values 1 and 2 only.
06 · Inclusive endpoint
For Z above, find P(1<Z≤4).
Hint
Include 2 and 4 but not 1.
Worked solution
2/7+4/7=6/7.
07 · Missing support value
What is P(Z=3)?
Hint
Which branch includes 3?
Worked solution
0. A formula for selected values does not automatically extend to intermediate integers.
05 / A parameter can have a range
Normalisation may hold for many parameter values.
If masses are a,0.3,0.7−a, their sum is 1 for every a.
Nonnegativity then requires 0≤a≤0.7.
At an endpoint a mass can be zero. A table may list such a value, but it is then not in the positive-probability support.
08 · Test a proposed parameter
Is a=0.8 valid for those masses?
Hint
Check the final entry.
Worked solution
No. The final mass is −0.1. A total of 1 cannot repair a negative probability.
09 · Valid interval
Masses are 2t,0.4,0.6−2t. Find all valid t.
Hint
Check both expressions involving t are nonnegative.
Worked solution
t≥0 and t≤0.3, so 0≤t≤0.3. Their sum is already 1.
06 / Select supported values first
Strict and inclusive signs can change the answer.
Use this distribution for the next questions: values −3,0,2,6 have respective masses 0.1,0.2,0.4,0.3.
10 · Strict range
Find P(0<X<6).
Hint
Only 2 is strictly inside both bounds.
Worked solution
0.4.
11 · Include the endpoints
Find P(0≤X≤6).
Hint
Include 0,2,6.
Worked solution
0.2+0.4+0.3=0.9.
12 · Complement
Find P(X≠2).
Hint
Complement the mass at 2.
Worked solution
1−0.4=0.6.
07 / Reject an invalid model before calculating
Both tests protect the meaning of probability.
13 · Negative weights
Could choosing a normalising constant turn weights 1,−2,4 into probabilities?
Hint
Their total is 3, so the only normalising multiplier is 1/3.
Worked solution
No. Normalisation gives 1/3,−2/3,4/3, which are invalid. Nonnegative weights are required.
14 · Formula without a domain
Why is P(X=x)=x/10 incomplete as a distribution specification?
Hint
Which values of x are allowed?
Worked solution
The supported values must be given. For example, on {1,2,3,4} the masses sum to 1; on {1,2,3} they sum to 0.6. The same expression on different sets can have different validity.
08 / List, normalise, validate, select
Keep support and probability separate.
Make a table if a formula is hard to read. Sum all entries to find a constant, then check their signs. For an event, choose the supported values satisfying the condition and add their masses. Never extend a PMF formula beyond its stated domain.
Section 1 of 8 · Check both requirements