Questions & explanations
1. A bag contains 4 red marbles and 6 blue marbles. Two marbles are drawn without replacement. What is the probability that at least one marble is red, given that the first marble drawn is blue?
- 1/3
- 4/9
- 5/9
- 2/3
- 3/4
Answer: 4/9
Given first marble is blue, after drawing one blue, remaining: 4 red, 5 blue (total 9). We need probability that at least one marble is red in the two draws, but since first is already blue, 'at least one red' means the second marble is red. So P(second red | first blue) = number of red remaining / total remaining = 4/9. Thus answer is 4/9.
2. A jar contains 4 red marbles and 6 blue marbles. Two marbles are drawn at random. Which of the following is true?
- The probability that both are red is greater when drawing with replacement than without replacement.
- The probability that both are red is the same with or without replacement.
- The probability that both are red is greater when drawing without replacement than with replacement.
- The probability that both are blue is the same with or without replacement.
- The probability that both are blue is greater with replacement than without replacement.
Answer: The probability that both are red is greater when drawing with replacement than without replacement.
With replacement: P(both red) = (4/10)*(4/10)=16/100=0.16. Without replacement: P(both red) = (4/10)*(3/9)=12/90≈0.1333. So with replacement gives higher probability. For blue: with replacement (6/10)^2=0.36; without replacement (6/10)*(5/9)=30/90≈0.3333, so with replacement also higher. Thus only option a is correct.
3. In a certain game, a player draws two cards without replacement from a standard deck of 52 cards. If both cards are hearts, the player wins $100. If exactly one card is a heart, the player wins $10. If no hearts are drawn, the player loses $20. What is the expected value of the player's winnings? (Round to the nearest cent.)
- -$2.94
- -$1.47
- $0.00
- $1.47
- $2.94
Answer: -$1.47
The expected value is calculated as: P(both hearts) = (13/52)*(12/51) = 156/2652, win $100; P(exactly one heart) = 2*(13/52)*(39/51) = 1014/2652, win $10; P(no hearts) = (39/52)*(38/51) = 1482/2652, lose $20. EV = (156*100 + 1014*10 + 1482*(-20))/2652 = (15600 + 10140 - 29640)/2652 = (-3900)/2652 ≈ -$1.47.
4. A company has 10 employees: 6 in sales and 4 in marketing. Two employees are selected at random to form a committee. Which of the following best describes the events 'first employee is from sales' and 'second employee is from marketing'?
- Independent, because the probability of the second event does not depend on the first.
- Dependent, because the probability of the second event changes after the first selection.
- Independent, because the selections are random.
- Dependent, because the events are mutually exclusive.
- Independent, because the events are from different departments.
Answer: Dependent, because the probability of the second event changes after the first selection.
Without replacement, the probability that the second is from marketing depends on whether the first was from sales. If first is sales, P(second marketing)=4/9; if first is marketing, P(second marketing)=3/9. Thus events are dependent.
5. Events A and B have probabilities P(A)=0.6, P(B)=0.5, and P(A∪B)=0.8. Are A and B independent?
- Yes, because P(A∩B)=0.3
- Yes, because P(A|B)=0.6
- No, because P(A∩B)=0.3
- No, because P(A|B)=0.5
- Cannot be determined
Answer: Yes, because P(A∩B)=0.3
By inclusion-exclusion, P(A∩B)=P(A)+P(B)-P(A∪B)=0.6+0.5-0.8=0.3. For independence, need P(A∩B)=P(A)P(B)=0.6*0.5=0.3. Since 0.3=0.3, they are independent. Also P(A|B)=P(A∩B)/P(B)=0.3/0.5=0.6=P(A), confirming independence.
6. A dartboard consists of a square of side length 2 feet, with a circle of radius 1 foot inscribed in the square. A dart is thrown randomly and lands at a uniformly random point within the square. If the dart lands inside the circle, the player scores 10 points; if it lands outside the circle but inside the square, the player scores 2 points. What is the expected value of the score from one throw?
- 2 + 2π
- 2 + 4π
- 2 + 2π/3
- 2 + 4π/3
- 2 + 2π/5
Answer: 2 + 2π
Area of square = 4 ft². Area of circle = π(1)² = π ft². Probability inside circle = π/4, outside = (4-π)/4 = 1 - π/4. Expected score = (π/4)*10 + (1 - π/4)*2 = (10π/4) + 2 - (2π/4) = 2 + (8π/4) = 2 + 2π.
7. A bag contains 5 red marbles and 3 blue marbles. One marble is drawn at random. What is the probability that the marble is either red or blue?
- 1/8
- 5/8
- 3/8
- 1
- 0
Answer: 1
The events 'draw a red marble' and 'draw a blue marble' are mutually exclusive (cannot happen at the same time) and together cover all outcomes. So P(red or blue) = P(red) + P(blue) = 5/8 + 3/8 = 1.
8. A game consists of rolling a fair six-sided die once. If the outcome is 1 or 2, the player wins $5. If the outcome is 3 or 4, the player wins $10. If the outcome is 5 or 6, the player loses $8 (i.e., wins -$8). What is the expected value of the player's winnings per game?
- $1.00
- $2.33
- $3.00
- $4.00
- $5.00
Answer: $2.33
Probability of each outcome: 1/6. Expected value = (1/6)*5 + (1/6)*5 + (1/6)*10 + (1/6)*10 + (1/6)*(-8) + (1/6)*(-8) = (1/6)[5+5+10+10-8-8] = (1/6)[14] = 14/6 ≈ 2.3333. So $2.33.
9. A fair six-sided die is rolled twice. What is the probability that the first roll is a 3 and the second roll is an even number?
- 1/12
- 1/6
- 1/9
- 1/18
- 1/36
Answer: 1/12
The probability of rolling a 3 on a fair die is 1/6. The probability of rolling an even number (2,4,6) is 3/6 = 1/2. Since the rolls are independent, multiply: (1/6)*(1/2)=1/12.
10. In a survey of 100 people, 60 like coffee, 50 like tea, and 40 like juice. 25 like both coffee and tea, 20 like both coffee and juice, 15 like both tea and juice, and 10 like all three. How many people like at least one of the three beverages?
- 80
- 90
- 100
- 110
- 120
Answer: 100
Use inclusion-exclusion for three events: |A∪B∪C| = |A|+|B|+|C| - |A∩B| - |A∩C| - |B∩C| + |A∩B∩C| = 60+50+40 -25-20-15 +10 = 150 -60 +10 = 100. So 100 people like at least one.
11. A box contains 3 red balls and 2 green balls. Two balls are drawn sequentially without replacement. What is the probability that the second ball is green, given that the first ball was red?
- 1/5
- 1/4
- 1/3
- 1/2
- 2/5
Answer: 1/2
Given first ball is red, after drawing one red, remaining: 2 red, 2 green (total 4). Probability second ball is green = number of green remaining / total remaining = 2/4 = 1/2.
12. A committee of 3 people is to be selected from a group of 5 men and 4 women. What is the probability that the committee consists of exactly 2 men and 1 woman?
- 10/21
- 5/14
- 3/7
- 20/63
- 40/63
Answer: 10/21
Total ways to choose 3 from 9: C(9,3)=84. Favorable ways: choose 2 men from 5: C(5,2)=10, choose 1 woman from 4: C(4,1)=4, total 10*4=40. Probability = 40/84 = 10/21.