Questions & explanations
1. A standard deck of 52 playing cards is shuffled. What is the probability that a hand of 5 cards contains exactly 2 aces?
- \frac{36}{54145}
- \frac{108}{54145}
- \frac{216}{54145}
- \frac{432}{54145}
- \frac{864}{54145}
Answer: \frac{108}{54145}
Total number of 5-card hands: C(52,5) = 2,598,960. Number of hands with exactly 2 aces: choose 2 aces from 4: C(4,2)=6, and choose 3 non-aces from 48: C(48,3)=17,296. Product = 6*17,296 = 103,776. Probability = 103,776 / 2,598,960 = simplify dividing by 24: 4,324 / 108,290 = further divide by 2: 2,162 / 54,145 = 108/54145.
2. A game costs $5 to play. You roll a fair six-sided die. If you roll a 6, you win $20. If you roll a 5, you win $10. Otherwise, you win nothing. What is the expected value of playing the game?
- -$0.50
- $0.00
- $0.50
- $1.00
- $5.00
Answer: $0.00
The expected value is calculated as the sum of each outcome's net gain times its probability. Net gains: roll 6: $20 - $5 = $15; roll 5: $10 - $5 = $5; otherwise: -$5. Probabilities: 1/6 for 6, 1/6 for 5, 4/6 for others. EV = (1/6)*15 + (1/6)*5 + (4/6)*(-5) = (15+5-20)/6 = 0/6 = $0.00.
3. A fair coin is tossed 4 times. What is the probability of getting at least one head and exactly two tails?
- 3/8
- 1/2
- 5/8
- 11/16
- 1/4
Answer: 3/8
Exactly two tails in 4 tosses: number of sequences = C(4,2)=6. Each sequence probability = (1/2)^4 = 1/16. So P(exactly two tails) = 6/16 = 3/8. Since at least one head is automatically true if exactly two tails (because 4-2=2 heads), the condition is redundant. Thus answer = 3/8.
4. A diagnostic test for a disease has a 95% sensitivity (true positive rate) and 90% specificity (true negative rate). The disease prevalence is 1%. If a randomly selected person tests positive, what is the probability that they actually have the disease?
- 0.087
- 0.095
- 0.105
- 0.500
- 0.950
Answer: 0.087
Let D = has disease, T+ = positive test. P(D) = 0.01, P(T+|D) = 0.95, P(T+|not D) = 0.10. P(T+) = P(T+|D)P(D) + P(T+|not D)P(not D) = 0.95*0.01 + 0.10*0.99 = 0.0095 + 0.099 = 0.1085. P(D|T+) = (0.95*0.01)/0.1085 ≈ 0.0095/0.1085 ≈ 0.0876 ≈ 0.087.
5. A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn at random with replacement. What is the probability that both marbles are red?
- 25/64
- 5/8
- 25/56
- 5/16
- 15/64
Answer: 25/64
Since the draws are with replacement, the events are independent. P(red on first draw) = 5/8. P(red on second draw) = 5/8. By multiplication rule for independent events, P(both red) = (5/8)*(5/8) = 25/64.
6. In a certain city, 60% of residents own a car, 30% own a bicycle, and 20% own both. What is the probability that a randomly selected resident owns a car or a bicycle but not both?
- 0.50
- 0.70
- 0.90
- 0.10
- 0.30
Answer: 0.50
P(car only) = P(car) - P(both) = 0.60 - 0.20 = 0.40. P(bicycle only) = P(bicycle) - P(both) = 0.30 - 0.20 = 0.10. Since these are mutually exclusive, P(car or bicycle but not both) = 0.40 + 0.10 = 0.50.
7. A bag contains 5 red marbles and 3 blue marbles. If 3 marbles are drawn at random without replacement, what is the probability that exactly 2 are red?
- 15/28
- 5/14
- 10/21
- 5/7
- 3/7
Answer: 15/28
Total ways to choose 3 marbles from 8: C(8,3)=56. Ways to choose 2 red from 5: C(5,2)=10, and 1 blue from 3: C(3,1)=3. Favorable = 10*3=30. Probability = 30/56 = 15/28.
8. A game consists of rolling a fair 6-sided die. If the result is 1 or 2, you win $10. If the result is 3 or 4, you win $5. If the result is 5 or 6, you lose $8. What is the expected value of the game?
- $2.33
- $3.00
- $1.67
- $0.00
- $7.00
Answer: $2.33
P(1 or 2) = 1/3, win $10; P(3 or 4) = 1/3, win $5; P(5 or 6) = 1/3, lose $8 (i.e., -$8). Expected value = (1/3)*10 + (1/3)*5 + (1/3)*(-8) = (10+5-8)/3 = 7/3 ≈ $2.33.
9. A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn at random with replacement. What is the probability that both marbles are red?
- 25/64
- 5/8
- 5/16
- 25/56
- 5/14
Answer: 25/64
Since the draws are with replacement, the events are independent. P(red on first draw) = 5/8. P(red on second draw) = 5/8. Multiplication rule: (5/8)*(5/8) = 25/64.
10. A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn at random without replacement. What is the probability that exactly one of the two marbles is red?
- 15/28
- 15/56
- 30/56
- 5/14
- 3/7
Answer: 15/28
Probability exactly one red = P(red then blue) + P(blue then red). P(red then blue) = (5/8)*(3/7)=15/56. P(blue then red) = (3/8)*(5/7)=15/56. Sum = 30/56 = 15/28.
11. A diagnostic test for a disease has a 95% sensitivity (true positive rate) and 90% specificity (true negative rate). The disease prevalence is 2%. If a randomly selected person tests positive, what is the probability that they actually have the disease?
- 0.162
- 0.190
- 0.210
- 0.237
- 0.950
Answer: 0.162
Using Bayes' theorem: P(D|+) = P(+|D)*P(D) / [P(+|D)*P(D) + P(+|~D)*P(~D)] = (0.95*0.02) / (0.95*0.02 + 0.10*0.98) = 0.019 / (0.019 + 0.098) = 0.019/0.117 ≈ 0.162.
12. A disease affects 1% of the population. A test is 95% accurate: it correctly identifies 95% of those with the disease (true positive) and 95% of those without (true negative). If a person tests positive, what is the probability they actually have the disease?
- 0.161
- 0.950
- 0.500
- 0.050
- 0.0095
Answer: 0.161
Let D = disease, T+ = positive. P(D)=0.01, P(T+|D)=0.95, P(T+|not D)=0.05. P(T+) = 0.01*0.95 + 0.99*0.05 = 0.0095+0.0495=0.059. P(D|T+) = 0.0095/0.059 ≈ 0.161.