Learn with Amar
Teaching video
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Undergraduate chapters and video availability01 · Read and understand
What you will learn
- Derive memorylessness for independent repeated Bernoulli trials.
- Justify the conclusion "The probability is (3/4)³=27/64" using the stated assumptions.
Before you start
Geometric series and conditional probability.
Keep paper nearby. Read the question once for the context, then again to identify what is known and what you need to find.
Start with a question
If success probability is p=1/4 per independent attempt and X counts attempts through the first success, find P(X>5 | X>2).
Why this math matters
Derive memorylessness for independent repeated Bernoulli trials. This worked micro-lesson connects a precise mathematical condition to a conclusion you can check. The transfer task asks you to change the setting and decide which parts of the reasoning still apply.
Set up the model
A useful answer starts with clear assumptions:
- Attempts are independent.
- Success probability remains one quarter on every attempt.
02 · Work through the example
Follow the reasoning, one step at a time.
Try to predict the next step before reading it. After each calculation, explain why the operation makes sense and how it helps answer the original question.
Work through it with Amar
See this example unfold.
The complete worked example, one idea at a time.
Understand a waiting time that forgets failures
PausedQuestion: Start with the question. Paused.
Question
Start with the question
If success probability is p=1/4 per independent attempt and X counts attempts through the first success, find P(X>5 | X>2).
Before you calculate
Read what is known and what you need to find. Make a prediction before moving to the first calculation.
Starts paused. Play advances through the full text at a reading pace; pause whenever you need more time. Previous, Next, and the phase buttons let you set your own pace. Playback pauses when this walkthrough leaves the screen or you switch tabs.
Your device’s reduced-motion setting keeps each phase still. Manual controls remain available. The full written solution stays below.
Build the model
P(X>k)=(3/4)ᵏ
Waiting beyond k attempts requires k failures.
Work through the mathematics
P(X>5 | X>2)=(3/4)⁵/(3/4)²
The longer waiting event is contained in the shorter one.
Check the conclusion
The probability is (3/4)³=27/64
After two failures, the next three attempts behave like a fresh sequence.
The result
The probability is (3/4)³=27/64
After two failures, the next three attempts behave like a fresh sequence.
Common mistakes to catch
- State whether the variable counts attempts or failures.
- A long wait does not make the next success more likely in this model.
03 · Practice independently
Try it before revealing the answer.
Use paper or a calculator as needed. Write your units and reasoning, then open the hint or explanation to check your approach.
Practice 1
Find E[X].
Show a hint
Sum the geometric tail probabilities.
Reveal answer and explanation
Four
Σₖ≥0(3/4)ᵏ=1/(1−3/4)=4.
Practice 2
Does the property survive if success probability improves with practice?
Show a hint
Compare future trial probabilities before and after failures.
Reveal answer and explanation
Generally no
The identical-trial assumption no longer holds.
Take the idea with you
Identify when repeated service attempts do and do not justify a geometric model.
04 · Reflect and continue
Can you explain it in your own words?
Before moving on, explain the main idea without looking at the worked example. Try both practice questions, check your reasoning, and name one mistake you now know how to avoid. Return to a step if you still need support.
Up next: Combine independent streams of rare events
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