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05-02: Exercises — Probability Rules and Conditional Probability

Notes reference: 05-02: Probability Rules and Conditional Probability

Q1: Mutually exclusive or not?

State whether each pair is mutually exclusive, then compute P(A or B).

  1. A = draw a king, B = draw a queen
  2. A = draw a king, B = draw a heart
  3. A = roll an even number, B = roll a 5
  4. A = roll an even number, B = roll a number > 3

Solution

1. MUTUALLY EXCLUSIVE (a card cannot be both)
   P(K or Q) = 4/52 + 4/52 = 8/52 = 0.1538

2. NOT mutually exclusive — the king of hearts is both
   P(K or ♥) = 4/52 + 13/52 − 1/52 = 16/52 = 0.3077

3. MUTUALLY EXCLUSIVE (5 is odd)
   P(even or 5) = 3/6 + 1/6 = 4/6 = 0.6667

4. NOT mutually exclusive — 4 and 6 are in both
   P(even or >3) = 3/6 + 3/6 − 2/6 = 4/6 = 0.6667

The tell: if the events can happen together, you must subtract the overlap.

Q2: With and without replacement

An urn holds 6 red and 4 blue marbles. Draw two.

Find P(both red): (a) with replacement, (b) without replacement.

Solution

(a) WITH replacement — INDEPENDENT
    P(R₁ and R₂) = P(R₁) × P(R₂) = (6/10)(6/10) = 36/100 = 0.3600

(b) WITHOUT replacement — DEPENDENT
    P(R₁ and R₂) = P(R₁) × P(R₂ | R₁) = (6/10)(5/9) = 30/90 = 0.3333

The second draw's probability changed from 6/10 to 5/9 because one red
marble is gone. That is exactly what "dependent" means.

Also find P(one of each colour) without replacement:

P(R then B) + P(B then R) = (6/10)(4/9) + (4/10)(6/9)
                          = 24/90 + 24/90 = 48/90 = 0.5333

Q3: Conditional probability from a table

A company surveys 400 employees:

Satisfied Neutral Dissatisfied Total
Remote 110 40 20 170
Hybrid 85 35 30 150
On-site 30 25 25 80
Total 225 100 75 400

Find:

  1. P(Satisfied)
  2. P(Remote)
  3. P(Remote and Satisfied)
  4. P(Satisfied | Remote)
  5. P(Remote | Satisfied)
  6. P(Remote or Satisfied)
  7. Are work mode and satisfaction independent?

Solution

1.  P(Satisfied)              = 225/400 = 0.5625          (marginal)
2.  P(Remote)                 = 170/400 = 0.4250          (marginal)
3.  P(Remote and Satisfied)   = 110/400 = 0.2750          (joint)

4.  P(Satisfied | Remote)     = 110/170 = 0.6471   ← divide by the ROW total
5.  P(Remote | Satisfied)     = 110/225 = 0.4889   ← divide by the COLUMN total

6.  P(Remote or Satisfied)    = 0.4250 + 0.5625 − 0.2750 = 0.7125
                                (or directly: (170 + 225 − 110)/400 = 285/400)

7.  INDEPENDENCE CHECK
    P(Satisfied | Remote) = 0.6471
    P(Satisfied)          = 0.5625
    0.6471 ≠ 0.5625  →  DEPENDENT

    Equivalently, the expected count if independent would be
        170 × 225 / 400 = 95.6,  but 110 was observed.
    That gap is what the chi-square test of 12-02 formalises.

Note how 4 and 5 differ. P(A|B) and P(B|A) are different questions with different denominators — 0.647 vs. 0.489.

Q4: Build the table in software

Reproduce the Q3 probabilities from raw data.

Solution

' Counts in B2:D4, totals in E and row 5
=SUM(B2:D2)                     ' E2  row total
=SUM(B2:B4)                     ' B5  column total
=B2/$E$5                        ' joint        P(Remote and Satisfied) -> 0.275
=E2/$E$5                        ' marginal     P(Remote)               -> 0.425
=B2/$E2                         ' row-cond     P(Satisfied | Remote)   -> 0.6471
=B2/B$5                         ' col-cond     P(Remote | Satisfied)   -> 0.4889
=E2*B5/$E$5                     ' expected if independent              -> 95.625

' From raw data:  Insert ▸ PivotTable
'   Rows = mode, Columns = satisfaction, Values = Count
'   Show Values As ▸ % of Grand Total  → joint
'   Show Values As ▸ % of Row Total    → P(column | row)
'   Show Values As ▸ % of Column Total → P(row | column)

tab <- matrix(c(110,40,20, 85,35,30, 30,25,25), nrow = 3, byrow = TRUE,
              dimnames = list(mode = c("Remote","Hybrid","Onsite"),
                              sat  = c("Satisfied","Neutral","Dissatisfied")))
addmargins(tab)
prop.table(tab)                        # joint
margin.table(tab, 1) / sum(tab)        # marginal by row
prop.table(tab, 1)                     # P(sat | mode)   — rows sum to 1
prop.table(tab, 2)                     # P(mode | sat)   — cols sum to 1
outer(rowSums(tab), colSums(tab)) / sum(tab)     # expected if independent

tab = pd.DataFrame([[110,40,20],[85,35,30],[30,25,25]],
                   index=["Remote","Hybrid","Onsite"],
                   columns=["Satisfied","Neutral","Dissatisfied"])
n = tab.values.sum()
tab / n                                        # joint
tab.sum(axis=1) / n                            # marginal
tab.div(tab.sum(axis=1), axis=0)               # P(sat | mode)
tab.div(tab.sum(axis=0), axis=1)               # P(mode | sat)
np.outer(tab.sum(axis=1), tab.sum(axis=0)) / n # expected if independent

Q5: Tree diagram

A factory has two suppliers. Supplier A provides 70% of parts with a 3% defect rate; supplier B provides 30% with a 6% defect rate.

  1. Draw the tree and find all four joint probabilities.
  2. P(defective)
  3. P(good)

Solution

                  0.03  Defective       P = 0.70 × 0.03 = 0.021
        0.70  A ─┤
       ┌─────────┘ 0.97  Good           P = 0.70 × 0.97 = 0.679
Start ─┤
       └─────────┐ 0.06  Defective      P = 0.30 × 0.06 = 0.018
        0.30  B ─┤
                  0.94  Good            P = 0.30 × 0.94 = 0.282
                                        ───────────────────────
                                        Total            1.000  ✓

2.  P(defective) = 0.021 + 0.018 = 0.039        (3.9%)
3.  P(good)      = 0.679 + 0.282 = 0.961        (or 1 − 0.039 ✓)

Multiply along a branch, add across branch endings. Step 2 is the Law of Total Probability.

Q6: Independence, three ways

P(A) = 0.4, P(B) = 0.5, P(A and B) = 0.2.

  1. Are A and B independent? Show it three ways.
  2. Find P(A | B) and P(B | A).
  3. Find P(A or B).

Solution

1.  TEST 1:  P(A and B) = P(A)·P(B)?
             0.2  =?  0.4 × 0.5 = 0.2       ✓ YES

    TEST 2:  P(A | B) = P(A)?
             0.2/0.5 = 0.4  =?  0.4         ✓ YES

    TEST 3:  P(B | A) = P(B)?
             0.2/0.4 = 0.5  =?  0.5         ✓ YES

    A and B are INDEPENDENT.

2.  P(A | B) = 0.2/0.5 = 0.40
    P(B | A) = 0.2/0.4 = 0.50

3.  P(A or B) = 0.4 + 0.5 − 0.2 = 0.70

Q7: Mutually exclusive vs. independent

P(A) = 0.3, P(B) = 0.4, and A and B are mutually exclusive.

  1. Find P(A and B).
  2. Find P(A or B).
  3. Are they independent?

Solution

1.  Mutually exclusive  →  P(A and B) = 0

2.  P(A or B) = 0.3 + 0.4 − 0 = 0.70

3.  Independence requires P(A and B) = P(A)·P(B) = 0.3 × 0.4 = 0.12
    But P(A and B) = 0  ≠  0.12
    →  NOT independent. They are DEPENDENT.

Intuition: if A occurs, B definitely did NOT — knowing A tells you
everything about B. That is the opposite of independence.

GENERAL RULE: two events with non-zero probability that are mutually
exclusive are ALWAYS dependent. "Mutually exclusive" and "independent"
are not synonyms — they are close to opposites.

Q8: A three-stage problem

A student answers three multiple-choice questions by guessing. Each has 4 options.

  1. P(all three correct)
  2. P(none correct)
  3. P(at least one correct)
  4. P(exactly one correct)

Solution

Each question:  P(correct) = 0.25,  P(wrong) = 0.75,  independent

1.  P(all 3) = (0.25)³ = 0.015625              (1.6%)

2.  P(none)  = (0.75)³ = 0.421875              (42.2%)

3.  P(at least one) = 1 − P(none)
                    = 1 − 0.421875 = 0.578125  (57.8%)   ← complement rule

4.  Exactly one correct — THREE arrangements (CWW, WCW, WWC):
    P = 3 × (0.25)(0.75)(0.75) = 3 × 0.140625 = 0.421875   (42.2%)

=0.25^3                             ' 0.015625
=1-0.75^3                           ' 0.578125
=BINOM.DIST(1, 3, 0.25, FALSE)      ' exactly one -> 0.421875

The binomial machinery behind #4 is developed in 06-02.

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