2.6 Conditional Probability and Multiplication — Answers and Rubrics
These solutions explain the new practice tasks in this lesson. For simulation parts, a valid recorded run may differ from another learner’s run; the rubric checks the model, recorded work and interpretation.
Original generated practice for the Fall 2026 course. Return to the lesson and independent checks.
Password Flags and Weak Passwords: worked answer
| 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.
. Eighty percent of weak passwords are flagged. . About 66.7% of flagged passwords are weak. The first denominator is 40 weak accounts; the second is 48 flagged accounts.
| Evidence for credit | Points |
|---|---|
| Computes .80 and 2/3. | 1 |
| Shows conditional ratios and denominators 40 versus 48. | 1 |
| Interprets each in the correct reference group. | 1 |
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.
. Reversing the order gives . The second factor is conditional on the first event. The rule works for dependent events; no independent-event product is assumed.
| Evidence for credit | Points |
|---|---|
| Shows both correct products and joint probability .16. | 1 |
| Explains conditional direction and that general multiplication does not require independence. | 1 |