πŸ§ͺ CHAPTER 3 Β· QUESTIONS

🎲 3.1 – brand new basic probability
3.1.1 – Spinner (1‑5) + card suit

A probability experiment: spin a spinner with equal sections 1,2,3,4,5 and draw a card from a standard deck and note its suit (hearts, diamonds, clubs, spades). List the sample space (use tree mentally). How many outcomes?

Counting: 5 spinner outcomes Γ— 4 suits = 20 outcomes. Sample space = {1β™₯,1♦,1♣,1β™ , 2β™₯,...,5β™ }. (explicit list not needed)
3.1.2 – Simple event? picking a Jack or a heart?

From a standard deck, consider event C = β€œthe card is a Jack or a heart”. Is C a simple event? Explain.

No – it contains multiple outcomes: all 4 Jacks plus the hearts that are not Jack (12 hearts) β†’ total 4+12=16 outcomes. Simple events have exactly one outcome.
3.1.3 – Pizza combinations

A pizza shop offers: crust (thin, thick, stuffed) ; sauce (red, pesto, BBQ) ; cheese (mozzarella, vegan, blend) ; topping (pepperoni, mushrooms, onions, sausage). How many different pizzas can you make with exactly one choice from each category?

3 crust Γ— 3 sauce Γ— 3 cheese Γ— 4 toppings = 3Β·3Β·3Β·4 = 108 pizzas.
3.1.4 – Two fair dice, find P(sum = 7) and P(sum > 9)

Roll two six‑sided dice (one red, one blue). Use classical probability.

Total outcomes = 36.
Sum=7: {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} β†’ 6 β†’ 6/36=1/6β‰ˆ0.1667.
Sum>9: sum 10,11,12. (4,6),(5,5),(6,4),(5,6),(6,5),(6,6) β†’ 6 outcomes β†’ 6/36=1/6.
3.1.5 – Laptop brand preference (empirical)
Brandfrequency
Dell84
HP63
Apple97
Lenovo42
Other14

If one respondent is chosen at random, what is the empirical probability that they prefer Apple or Dell?

Total = 84+63+97+42+14 = 300. Apple+Dell freq = 97+84=181 β†’ P = 181/300 β‰ˆ 0.6033.
3.1.6 – Classify probability type

a) β€œThe chance that it will rain tomorrow is 30%” – meteorologist’s forecast.
b) β€œThe probability of drawing a red card from a deck is 1/2.”
c) β€œBased on records, the probability a patient tests positive for a certain disease is 0.02.”

a) subjective (educated forecast, not long‑run frequency).
b) classical (equally likely).
c) empirical (from historical data).
3.1.7 – Complement rule (flight delays)

At a certain airport, the probability that a flight is delayed is 0.18. Find the probability that a flight is not delayed.

P(not delayed) = 1 – P(delayed) = 1 – 0.18 = 0.82.
3.1.8 – Marble then coin

A bag contains 3 red marbles and 5 blue marbles. You draw one marble (without replacement, but it doesn't affect coin), then toss a fair coin. Use a tree to find P(red and tails).

P(red) = 3/8. P(tails) = 1/2. independent? marble draw does not affect coin β†’ multiply: (3/8)*(1/2) = 3/16 = 0.1875.
3.1.9 – License plate probability

A state license plate has 3 letters followed by 3 digits (letters A‑Z, digits 0‑9). Letters and digits can repeat. What is the probability that a randomly generated plate reads β€œABC 123”?

Total plates = 26Β³ Γ— 10Β³ = 17,576 Γ— 1,000 = 17,576,000. Only one specific plate β†’ P = 1 / 17,576,000 β‰ˆ 5.69e-8.
πŸ”— 3.2 – fresh conditional & multiplication
3.2.1 – Candies: two draws without replacement

A jar contains 6 strawberry candies and 4 orange candies. You draw two candies without replacement. Find P(second is orange | first is strawberry).

After removing one strawberry: left 5 strawberry + 4 orange = 9 total. P(orange|first strawberry) = 4/9 β‰ˆ 0.444.
3.2.2 – Smartphone vs age group
iPhoneAndroidOther
under 3045325
30 or older285416

Find P(iPhone | under 30).

Total under 30 = 45+32+5 = 82. iPhone under30 = 45 β†’ 45/82 β‰ˆ 0.549.
3.2.3 – Independent? marble with replacement

Draw a marble from a bag (3 red, 5 blue), note color, replace it, then draw again. Are the events β€œfirst red” and β€œsecond blue” independent?

Yes, because replacement restores original composition. P(second blue|first red) = P(blue) = 5/8 β†’ independent.
3.2.4 – P(king then queen) with replacement

Two cards drawn from a deck with replacement. Find P(king first and queen second).

P(king) = 4/52, P(queen) = 4/52, independent β†’ (4/52)*(4/52)=16/2704 = 1/169 β‰ˆ 0.00592.
3.2.5 – P(spin ≀2 and draw heart)

Spin spinner 1‑8 equally likely, draw a card from deck. P(spin ≀2 and heart).

P(spin ≀2)=2/8=1/4. P(heart)=13/52=1/4. independent β†’ (1/4)*(1/4)=1/16=0.0625.
3.2.6 – P(all 4 bulbs work)

Probability a certain LED bulb works for 5 years is 0.92. Four bulbs are installed independently. Find probability all four still work after 5 years.

(0.92)⁴ = 0.7164 (approx).
3.2.7 – P(at least one bulb fails)

From previous, find probability that at least one of the four bulbs fails.

P(at least one fails) = 1 – P(all work) = 1 – 0.7164 = 0.2836.
3.2.8 – Disease testing

A test for a disease has sensitivity 0.95 (P(positive | disease)). Prevalence of disease is 0.02. What is P(disease and positive)?

P(disease and positive) = P(disease)*P(positive|disease) = 0.02 * 0.95 = 0.019.
3.2.9 – P(regular exercise | female)
MaleFemale
regular exercise3847
no regular exercise4233

Find P(regular exercise | female).

Total female = 47+33=80. Regular female = 47 β†’ 47/80 = 0.5875.
βž• 3.3 – original addition rule problems
3.3.1 – mutually exclusive? die: odd / greater than 4

Roll a die. A = odd, B = greater than 4. Are they mutually exclusive?

Outcomes A = {1,3,5}, B = {5,6}. Intersection = {5} not empty β†’ not mutually exclusive.
3.3.2 – male vs honors student mutually exclusive?

At a school: event M = male, event H = honors student. Can they be mutually exclusive?

No, a male can also be honors β†’ they are not mutually exclusive (unless no male honors exist, but in general they aren’t).
3.3.3 – P(heart or spade) from deck
Heart and spade are mutually exclusive (no card is both). P(heart)=13/52, P(spade)=13/52 β†’ sum = 26/52 = 0.5.
3.3.4 – P(face card or red)

Standard deck. Find P(face card or red).

P(face)=12/52, P(red)=26/52, P(face and red)=6/52 (red face cards: J♦,Q♦,K♦,Jβ™₯,Qβ™₯,Kβ™₯) β†’ 12/52+26/52-6/52 = 32/52 β‰ˆ 0.6154.
3.3.5 – Bus arrival (mutually exclusive)
minutes latefrequency
0-414
5-927
10-1422
15-199

Probability next bus is 0‑4 min late OR 10‑14 min late?

Total = 14+27+22+9 = 72. P(0‑4)=14/72, P(10‑14)=22/72, mutually exclusive β†’ (14+22)/72 = 36/72 = 0.5.
3.3.6 – P(cat or dog) from pet survey
DogNo dog
Cat3241
No cat5324

Find P(cat or dog).

Total = 32+41+53+24 = 150. Cat total = 32+41=73, dog total = 32+53=85, both = 32. P(cat or dog) = 73/150 + 85/150 – 32/150 = (73+85-32)/150 = 126/150 = 0.84.
3.3.7 – P(type A or Rh+)
ABABO
Rh+4218750
Rh-8429
Total = 42+18+7+50+8+4+2+9 = 140. P(A) = (42+8)/140=50/140, P(Rh+) = (42+18+7+50)/140 =117/140, P(A and Rh+) = 42/140. Addition: 50/140 + 117/140 – 42/140 = 125/140 β‰ˆ 0.8929.

ACADEZI 2026