01 · At least one
For this fixed two-check experiment, find P(X≥1).
Hint
Complement X=0.
Worked solution
1−0.18=0.82.
Understand · explore · practise
Constructing probability distributions
Build a discrete distribution from an experiment, distinguish a success count from attempts used, and handle the final probability in a capped stopping rule.
Before you startProbability trees, independent trials and probability mass functions.
01 / Define the recorded quantity precisely
The same experiment can support several different variables.
“Number of successes” and “number of attempts made” need not have the same support or distribution.
Write the stopping rule before listing possible values.
In a fixed two-attempt experiment, a success count can be 0,1 or 2. In a process that stops on success, the number of attempts used starts at 1. A final failure may end the experiment because the allowed attempts are exhausted.
Change the stopping capExplore
An independent connection attempt succeeds with probability 0.25. Stop at the first success or after m attempts, whichever happens first. T is the number of attempts actually made.
With a cap of 1, exactly one attempt is made, whether it succeeds or fails.
02 / Group complete outcomes
The distribution combines paths with the same recorded result.
Two independent checks succeed with probabilities 0.4 and0.7. Both are always run. X is the success count.Worked example
SS: 0.4×0.7=0.28; SF: 0.4×0.3=0.12
The first pair gives X=2, the second X=1.
FS: 0.6×0.7=0.42; FF: 0.6×0.3=0.18
The first gives X=1, the second X=0.
P(X=0)=0.18, P(X=1)=0.54, P(X=2)=0.28
The masses sum to 1. The stage success probabilities differ.
02 · Exactly one
Why is P(X=1) not simply 0.4×0.3?
Hint
There is another order with one success.
Worked solution
That product only counts SF. Add FS as well: 0.12+0.42=0.54.
03 / Stop on the first success
An early stop requires failures followed by a success.
Independent connection attempts succeed with probability 1/4. Stop on success or after 4 attempts. T counts attempts made.Worked example
T=1: success immediately
P(T=1)=1/4.
T=2: fail, then succeed
P(T=2)=(3/4)(1/4)=3/16.
T=3: fail twice, then succeed
P(T=3)=(3/4)²(1/4)=9/64.
03 · Support
List the support of T.
Hint
An attempt is always made, and no more than four are allowed.
Worked solution
{1,2,3,4}.
04 · Why not0?
Can T=0 if every attempt fails?
Hint
T records attempts, not successes.
Worked solution
No. Four attempts are made in that case, so T=4.
04 / Treat the last value differently
Reaching the final attempt already guarantees T equals the cap.
Watch: combine both final outcomes
Pause, replay or seek freely. The notes explain the same idea and stay in view.
At the cap of 4:Worked example
To reach attempt 4, the first three must fail
This has probability (3/4)³=27/64.
The fourth attempt can succeed or fail
FFFS has probability 27/256; FFFF has probability 81/256.
Both paths give T=4
P(T=4)=(27+81)/256=27/64. Do not multiply the final mass by 1/4.
05 · Total check
Show that all four T probabilities sum to 1.
Hint
Use a common denominator 64.
Worked solution
16/64+12/64+9/64+27/64=1.
06 · First success versus final attempt
Is P(T=4) the probability of first success on attempt 4?
Hint
T=4 also allows failure on every attempt.
Worked solution
No. First success on attempt 4 has probability 27/256; T=4 has probability 27/64.
05 / Translate events back to the stopping rule
More than two attempts means the first two failed.
07 · More than two
Find P(T>2).
Hint
Fail twice; later outcomes do not affect this event.
Worked solution
(3/4)²=9/16. Also 9/64+27/64=36/64=9/16.
08 · At most two
Find P(T≤2).
Hint
Complement T>2.
Worked solution
1−9/16=7/16, also 1/4+3/16.
09 · Eventual success within the allowance
Find the chance of a success within the four attempts.
Hint
Complement four failures.
Worked solution
1−(3/4)⁴=175/256. This is different from P(T≤4)=1.
06 / Generalise carefully
The formula changes at the capped endpoint.
For cap m and independent success chance p:
P(T=t)=(1−p)^(t−1)p for 1≤t<m; P(T=m)=(1−p)^(m−1).
For m=1, T=1 certainly. For 0<p<1, all integers 1 through m have positive mass. At p=1 only T=1 occurs; at p=0 only T=m occurs. The cap makes this a different distribution from an unlimited waiting time.
10 · Cap of three
For p=1/4 and cap 3, list the masses.
Hint
The final value includes all failures before attempt 3.
Worked solution
T=1,2,3 have masses 1/4,3/16,9/16.
11 · Cap of five
For p=1/4 and cap 5, find P(T=5).
Hint
Only the first four failures are needed.
Worked solution
(3/4)⁴=81/256.
07 / Compare the two recorded quantities
A stopped success count is either zero or one.
12 · Success count in the capped process
Let S count successes before stopping, with p=1/4 and cap 4. Give its distribution.
Hint
There can be at most one success, because it ends the process.
Worked solution
P(S=0)=81/256 and P(S=1)=175/256. S differs from the attempts-used variable T.
13 · Unequal attempt chances
If attempt success probabilities change after a failure, can you use powers of 3/4?
Hint
The identical-probability assumption no longer applies.
Worked solution
Not without justification. Use the appropriate conditional failure probabilities along each path instead.
14 · A constant cap
If the experiment always makes exactly 4 attempts regardless of outcomes, what is the distribution of attempts used?
Hint
The stopping rule has changed.
Worked solution
The attempts-used variable is 4 with probability 1. The number of successes remains random.
08 / Build the table from the rule
Check what makes every supported value occur.
List paths, group those giving the same variable value, then add their probabilities. For a capped first-success process, early masses end in success; the last mass includes every path that reaches the final attempt. Check support, nonnegativity and a total of 1.
Section 1 of 8 · Define the recorded quantity precisely