Conditional Probability and Independence

Draw one card from a standard deck.

A = card is a Queen B = card is a face card (J, Q, K)

What’s P(Queen | Face card)?

Without condition: P(Queen) = 4/52 = 1/13

With condition: Given it’s a face card (12 cards), what’s P(Queen)? $P(A|B) = \frac{P(\text{Queen and Face})}{P(\text{Face})} = \frac{4/52}{12/52} = \frac{4}{12} = \frac{1}{3}$

Or directly: 4 queens among 12 face cards = 1/3

A bag has 5 red and 3 blue marbles. Draw 2 without replacement.

P(both red)?

A = first marble is red B = second marble is red

$P(A) = 5/8$ (5 red out of 8 total) $P(B|A) = 4/7$ (after removing one red: 4 red out of 7)

$P(A \cap B) = P(A) \times P(B|A) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$

Flip a fair coin twice.

A = first flip is heads B = second flip is heads

Are these independent?

$P(A) = 0.5$ $P(B|A) = 0.5$ (second flip doesn’t “know” about the first)

Since $P(B|A) = P(B)$, the events are independent.

$P(\text{both heads}) = P(A) \times P(B) = 0.5 \times 0.5 = 0.25$

A bag has 6 red and 4 green balls.

A = first draw is red B = second draw is red

$P(A) = 6/10 = 0.6$ $P(B|A) = 5/9 \approx 0.556$ (only 5 red left among 9 balls) $P(B|\text{not }A) = 6/9 \approx 0.667$ (all 6 red still among 9 balls)

Since $P(B|A) \neq P(B|\text{not }A)$, these events are dependent.

Survey of 200 people:

Are “morning person” and “coffee drinker” independent?

$P(\text{Coffee}) = 110/200 = 0.55$

$P(\text{Coffee}|\text{Morning}) = 60/100 = 0.60$

Since $0.60 \neq 0.55$, these events are dependent.

Morning people are more likely to drink coffee than night owls (60% vs 50%).

Draw 3 cards from a deck without replacement. P(all hearts)?

$P(\text{3 hearts}) = \frac{13}{52} \times \frac{12}{51} \times \frac{11}{50}$

$= \frac{13 \times 12 \times 11}{52 \times 51 \times 50} = \frac{1716}{132600} \approx 0.013$

About 1.3% chance of drawing 3 hearts in a row.

A disease affects 1% of the population. A test is:

• 95% accurate for people WITH disease (sensitivity)
• 90% accurate for people WITHOUT disease (specificity)

Tree structure:

Population
├── Disease (0.01)
│   ├── Test + (0.95) → True Positive
│   └── Test - (0.05) → False Negative
└── No Disease (0.99)
    ├── Test + (0.10) → False Positive
    └── Test - (0.90) → True Negative

P(Test positive)? $P(+) = P(D) \times P(+|D) + P(\bar{D}) \times P(+|\bar{D})$ $= (0.01)(0.95) + (0.99)(0.10)$ $= 0.0095 + 0.099 = 0.1085$

About 10.85% of all people test positive!

Factory has 3 machines producing widgets:

• Machine A: 50% of production, 2% defect rate
• Machine B: 30% of production, 3% defect rate
• Machine C: 20% of production, 5% defect rate

What’s P(randomly selected widget is defective)?

$P(D) = P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C)$ $= (0.50)(0.02) + (0.30)(0.03) + (0.20)(0.05)$ $= 0.01 + 0.009 + 0.01 = 0.029$

2.9% of all widgets are defective.

A = It’s raining B = Ground is wet

$P(\text{wet}|\text{rain}) \approx 0.99$ (rain almost always makes ground wet) $P(\text{rain}|\text{wet}) \approx 0.40$ (ground might be wet from sprinklers, hose, etc.)

These are very different probabilities!