2.6 Conditional Probability and Multiplication

“Given” Changes the Reference Group

Suppose 30% of shipments use Express service. Under this new model, 10% of Express shipments arrive late, while 20% of Standard shipments arrive late. Those last two percentages are conditional probabilities. The service has already been specified; the reference group is that service’s shipments.

Write $E$ for Express and $L$ for late. Then $P(L\mid E)=0.10$. The bar means “given,” so this reads “the probability of late given Express.” It does not mean $P(E\mid L)$. That reverse question starts with late shipments and asks which used Express.

For an event $B$ of positive probability,

Only the outcomes in $B$ remain in the denominator. The numerator is the overlap: outcomes that also satisfy $A$.

Worked Example: Follow Both Branches

The general multiplication rule says

It does not require independence. The second factor is the probability after the first event is known. For Express and late,

For Standard and late, the corresponding product is $0.70(0.20)=0.14$. The complete model can be organized as four paths:

Service	Service probability	Conditional status probability	Joint path probability
Express	0.30	Late: 0.10	0.03
Express	0.30	On time: 0.90	0.27
Standard	0.70	Late: 0.20	0.14
Standard	0.70	On time: 0.80	0.56

All paths sum to 1. The late paths give $P(L)=0.03+0.14=0.17$. Therefore

About 17.65% of late shipments use Express, while 10% of Express shipments are late. Both statements are consistent: they condition on different populations. It is often useful to imagine 10,000 shipments following the model: 300 Express-late, 2,700 Express-on-time, 1,400 Standard-late and 5,600 Standard-on-time. Then the reverse denominator is the 1,700 late shipments.

A tree diagram would multiply along each service/status path and add separate paths belonging to the same event. A table shows the same joint probabilities in cells. Whichever representation you use, keep the conditioning direction visible and check that the reference group has positive probability before dividing.

Try It Yourself

The following original practice applies the ideas above to a fresh setting. Show the reasoning requested in each part. Answers and task-specific rubrics are on the separate solutions page.

Original generated practice for the Fall 2026 course.

Password Flags and Weak Passwords

Complete population of 200 accounts

Password status	Flagged F	Not flagged	Total
Weak W	32	8	40
Not weak	16	144	160
Total	48	152	200

One account is selected uniformly from this complete population. Each has one weakness label and one flag label. Treat the table as exact population counts.

Part a

Calculate P(F given W) and P(W given F) using the conditional-probability formula. Identify the reference group and denominator for each and interpret both results.

Your reasoning

Part b

Use the general multiplication rule twice to calculate P(W and F), first starting with W and then starting with F. Explain why the second factor changes and why independence is not required.

Your reasoning

Worked answer and rubric