Multiple Choice Questions
Advanced Probability and Statistics (Continued)
Topic: Advanced Probability and Statistics
Grade: 12
Question 1:
A box contains 5 red balls, 3 blue balls, and 2 green balls. Two balls are drawn at random without replacement. What is the probability that both balls drawn are red?
a) 1/33
b) 5/33
c) 10/33
d) 1/2
Answer: b) 5/33
Explanation: The probability of drawing the first red ball is 5/10. After drawing the first red ball, there are 4 red balls left out of 9 total balls. So, the probability of drawing the second red ball is 4/9. Therefore, the probability of drawing both balls red is (5/10) * (4/9) = 20/90 = 5/33. This can also be thought of as a conditional probability problem, where the probability of drawing the second red ball is dependent on the first red ball being drawn.
Example:
Simple Example: If there are 2 red balls and 2 blue balls in a box, and two balls are drawn at random without replacement, the probability of drawing both balls red is 1/3.
Complex Example: If there are 10 red balls, 6 blue balls, and 4 green balls in a box, and three balls are drawn at random without replacement, the probability of drawing all three balls red is (10/20) * (9/19) * (8/18) = 120/684 = 10/57.
Question 2:
In a group of 50 students, 25 play football, 20 play basketball, and 10 play both sports. What is the probability that a randomly selected student plays either football or basketball?
a) 1/2
b) 3/5
c) 4/5
d) 9/10
Answer: c) 4/5
Explanation: The total number of students who play either football or basketball is 25 + 20 – 10 = 35. Therefore, the probability of selecting a student who plays either sport is 35/50 = 7/10. This can also be thought of as the union of two events, where the probability of selecting a student who plays football or basketball is equal to the sum of their individual probabilities minus the probability of their intersection.
Example:
Simple Example: If there are 30 students in a class and 15 play football, 10 play basketball, and 5 play both sports, the probability of selecting a student who plays either sport is 20/30 = 2/3.
Complex Example: If there are 100 students in a school and 60 play football, 50 play basketball, and 30 play both sports, the probability of selecting a student who plays either sport is 80/100 = 4/5.
Question 3:
The heights of a group of students are normally distributed with a mean of 65 inches and a standard deviation of 3 inches. What is the probability that a randomly selected student is taller than 70 inches?
a) 0.1587
b) 0.3085
c) 0.8413
d) 0.6915
Answer: a) 0.1587
Explanation: To find the probability of a randomly selected student being taller than 70 inches, we need to calculate the z-score. The z-score is calculated as (x – mean) / standard deviation. In this case, the z-score is (70 – 65) / 3 = 5/3. Using a standard normal distribution table, we can find that the probability corresponding to a z-score of 5/3 is approximately 0.0918. Since we are interested in the probability of being taller than 70 inches, we subtract this value from 0.5 to get 0.5 – 0.0918 = 0.1587.
Example:
Simple Example: If the heights of a group of students are normally distributed with a mean of 60 inches and a standard deviation of 5 inches, the probability of selecting a student taller than 65 inches is approximately 0.1587.
Complex Example: If the heights of a group of students are normally distributed with a mean of 70 inches and a standard deviation of 2 inches, the probability of selecting a student taller than 75 inches is approximately 0.0228.
Question 4:
A box contains 10 red balls and 8 blue balls. Three balls are drawn at random without replacement. What is the probability that all three balls drawn are red?
a) 1/136
b) 5/136
c) 15/136
d) 1/24
Answer: c) 15/136
Explanation: The probability of drawing the first red ball is 10/18. After drawing the first red ball, there are 9 red balls left out of 17 total balls. So, the probability of drawing the second red ball is 9/17. Similarly, the probability of drawing the third red ball is 8/16. Therefore, the probability of drawing all three balls red is (10/18) * (9/17) * (8/16) = 15/136.
Example:
Simple Example: If there are 5 red balls and 5 blue balls in a box, and three balls are drawn at random without replacement, the probability of drawing all three balls red is 1/42.
Complex Example: If there are 15 red balls and 10 blue balls in a box, and four balls are drawn at random without replacement, the probability of drawing all four balls red is (15/25) * (14/24) * (13/23) * (12/22) = 273/885.
Question 5:
A coin is flipped 10 times. What is the probability of getting exactly 7 heads?
a) 0.1172
b) 0.2051
c) 0.2813
d) 0.3828
Answer: a) 0.1172
Explanation: The probability of getting exactly 7 heads in 10 coin flips can be calculated using the binomial probability formula: P(X=k) = C(n,k) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successes, p is the probability of success on a single trial, and C(n,k) is the number of combinations. In this case, n=10, k=7, p=0.5, and C(10,7) = 10! / (7!(10-7)!) = 120. Plugging these values into the formula gives P(X=7) = 120 * (0.5)^7 * (1-0.5)^(10-7) = 0.1172.
Example:
Simple Example: If a coin is flipped 5 times, the probability of getting exactly 3 heads is 0.3125.
Complex Example: If a coin is flipped 15 times, the probability of getting exactly 10 heads is 0.1367.