Which table could be the probability distribution of a discrete random variable X that takes the values 0, 1 and 2?
P(0) = 0.2, P(1) = 0.5, P(2) = 0.3
✓
P(0) = 0.4, P(1) = 0.5, P(2) = 0.3
P(0) = 0.6, P(1) = 0.5, P(2) = -0.1
P(0) = 0.1, P(1) = 0.2, P(2) = 0.3
Correct Answer
P(0) = 0.2, P(1) = 0.5, P(2) = 0.3
A valid discrete distribution needs every probability between 0 and 1 and a total of exactly 1, and 0.2 + 0.5 + 0.3 = 1. The table P(0) = 0.4, P(1) = 0.5, P(2) = 0.3 adds to 1.2. The table P(0) = 0.6, P(1) = 0.5, P(2) = -0.1 adds to 1 but contains a negative probability. The table P(0) = 0.1, P(1) = 0.2, P(2) = 0.3 adds to only 0.6.
Category: Probability Distributions
Question 2
A discrete random variable X takes the values 1, 2, 3 and 4 with P(1) = 0.1, P(2) = 0.3 and P(3) = 0.4. What is P(4)?
0.1
0.2
✓
0.4
0.8
Correct Answer
0.2
The probabilities of a discrete distribution must add to 1, so P(4) = 1 - (0.1 + 0.3 + 0.4) = 1 - 0.8 = 0.2. The value 0.1 repeats P(1). The value 0.4 repeats P(3). The value 0.8 is the sum of the three given probabilities, not what remains.
Category: Probability Distributions
Question 3
A discrete random variable X takes the values 0, 1 and 2 with probabilities 0.3, 0.5 and 0.2. What is the expected value of X?
1
3
0.9
✓
0.5
Correct Answer
0.9
The expected value is the sum of each value times its probability: 0(0.3) + 1(0.5) + 2(0.2) = 0 + 0.5 + 0.4 = 0.9. The value 1 is the plain average of 0, 1 and 2, which ignores the probabilities. The value 3 is the sum of the three values. The value 0.5 is the largest single probability, not a mean.
Category: Mean and Standard Deviation
Question 4
A random variable X equals 0 with probability 0.5 and 4 with probability 0.5. What is the standard deviation of X?
4
16
1
2
✓
Correct Answer
2
The mean is 0(0.5) + 4(0.5) = 2. The variance is (0 - 2)^2(0.5) + (4 - 2)^2(0.5) = 2 + 2 = 4, so the standard deviation is sqrt(4) = 2. The value 4 is the variance, not its square root. The value 16 squares the variance. The value 1 does not match either the variance or its square root.
Category: Mean and Standard Deviation
Question 5
A game costs $2 to play. With probability 0.1 the player receives a $10 prize (the $2 is not returned), and otherwise receives nothing. What is the player's expected net gain per play, counting the cost?
-$1.00
✓
$1.00
-$0.80
$0.80
Correct Answer
-$1.00
A winner nets $10 - $2 = $8 with probability 0.1, and a loser nets -$2 with probability 0.9, so the expected net gain is 8(0.1) + (-2)(0.9) = 0.8 - 1.8 = -$1.00 per play; equivalently, the expected prize 0.1 x $10 = $1.00 minus the $2 cost. The value $1.00 is the expected prize before the $2 cost is subtracted. The value -$0.80 charges the $2 only on losing plays, as if a winner got the cost back. The value $0.80 counts only the winning outcome.
Category: Mean and Standard Deviation
Question 6
What does the expected value of a discrete random variable describe?
The value that X will take on the next trial
The long-run average of X over many repetitions
✓
The single most likely value of X
The middle value of the possible values of X
Correct Answer
The long-run average of X over many repetitions
The expected value, mu, is the average value of X that would be observed over a very large number of repetitions, weighting each value by its probability. The value that X will take on the next trial cannot be predicted, since X is random. The single most likely value of X is the mode, which may differ from the mean. The middle value of the possible values ignores how likely each one is.
Category: Mean and Standard Deviation
Question 7
Which random variable has a binomial distribution?
The number of flips of a fair coin until the first head
The number of aces in 5 cards dealt from one deck
The number of heads in 10 flips of a fair coin
✓
The number of emails that arrive in one hour
Correct Answer
The number of heads in 10 flips of a fair coin
Ten coin flips are a fixed number of independent trials, each with two outcomes and the same probability of heads, so the count of heads is binomial with n = 10 and p = 0.5. The number of flips until the first head has no fixed number of trials and is geometric. The number of aces in 5 cards is hypergeometric, because cards are dealt without replacement. The number of emails in one hour counts events in an interval of time, which fits a Poisson model.
Category: Binomial Distribution
Question 8
X has a binomial distribution with n = 20 and p = 0.3. What is the mean of X?
14
4.2
0.3
6
✓
Correct Answer
6
The mean of a binomial distribution is mu = np = 20(0.3) = 6. The value 14 is nq = 20(0.7), the expected number of failures. The value 4.2 is the variance npq = 20(0.3)(0.7). The value 0.3 is the probability of success on one trial.
Category: Binomial Distribution
Question 9
A basketball player makes each free throw with probability 0.5, independently, and takes 100 free throws. What is the standard deviation of the number of free throws made?
5
✓
25
50
10
Correct Answer
5
The standard deviation of a binomial distribution is sqrt(npq) = sqrt(100 x 0.5 x 0.5) = sqrt(25) = 5. The value 25 is the variance npq, before taking the square root. The value 50 is the mean np. The value 10 is the square root of n alone.
Category: Binomial Distribution
Question 10
A multiple choice quiz has 3 questions, and a student guesses on each with probability 0.2 of being right. Answers are independent. What is the probability the student gets exactly 1 question right?
0.128
0.384
✓
0.2
0.512
Correct Answer
0.384
This is binomial with n = 3 and p = 0.2, so P(X = 1) = 3 x 0.2 x 0.8^2 = 3 x 0.2 x 0.64 = 0.384. The factor 3 counts the three positions where the single right answer can fall. The value 0.128 leaves out that factor of 3. The value 0.2 is the probability of a right answer on one question. The value 0.512 is 0.8^3, the probability of getting none right.
Category: Binomial Distribution
Question 11
X has a binomial distribution with n = 4 and p = 0.1. What is P(X = 0), rounded to four decimal places?
0.0001
0.3439
0.6561
✓
0.9000
Correct Answer
0.6561
X = 0 means all 4 trials are failures, each with probability 0.9, so P(X = 0) = 0.9^4 = 0.6561. The value 0.0001 is 0.1^4, the probability that all 4 trials are successes. The value 0.3439 is 1 - 0.6561, the probability of at least one success. The value 0.9000 is the probability of failure on a single trial.
Category: Binomial Distribution
Question 12
Each attempt independently succeeds with probability 0.2. X is the number of attempts up to and including the first success. What is the mean of X?
4
0.2
20
5
✓
Correct Answer
5
For a geometric random variable that counts trials up to and including the first success, mu = 1/p = 1/0.2 = 5. The value 4 is (1 - p)/p, the mean number of failures before the first success, a different random variable. The value 0.2 is the probability of success itself. The value 20 is the variance (1 - p)/p^2 = 0.8/0.04, not the mean.
Category: Geometric Distribution
Question 13
Each attempt independently succeeds with probability 0.25. What is the probability that the first success happens on the third attempt, rounded to four decimal places?
0.1406
✓
0.0156
0.4219
0.1875
Correct Answer
0.1406
The first two attempts must fail and the third must succeed, so the probability is (0.75)^2(0.25) = 0.5625 x 0.25 = 0.140625, which rounds to 0.1406. The value 0.0156 is 0.25^3, three successes in a row. The value 0.4219 is 0.75^3, three failures in a row. The value 0.1875 is 0.75 x 0.25, which allows only one failure before the success.
Category: Geometric Distribution
Question 14
Which feature separates a hypergeometric experiment from a binomial experiment?
The trials are independent
Items are sampled without replacement
✓
The count of trials is not fixed
Events are counted over a time interval
Correct Answer
Items are sampled without replacement
A hypergeometric experiment draws a sample without replacement from a population made of two groups, so each pick changes the chances for later picks. Trials that are independent describe a binomial experiment, not a hypergeometric one. A count of trials that is not fixed describes a geometric experiment. Events counted over a time interval describe a Poisson experiment.
Category: Hypergeometric Distribution
Question 15
A committee of 5 is chosen at random from a group of 6 women and 4 men. X is the number of women on the committee. What is the mean of X?
2
5
3
✓
6
Correct Answer
3
X is hypergeometric with a group of interest of 6 women, a second group of 4 men and a sample of 5, so mu = n x r / (r + b) = 5 x 6 / 10 = 3. The value 2 is the mean number of men, 5 x 4 / 10. The value 5 is the committee size. The value 6 is the number of women available, not the number expected on the committee.
Category: Hypergeometric Distribution
Question 16
A box holds 4 red and 6 blue chips. Two chips are drawn at random without replacement. What is the probability that exactly 1 chip is red?
4/15
12/25
2/5
8/15
✓
Correct Answer
8/15
This is hypergeometric: P(X = 1) = C(4,1) x C(6,1) / C(10,2) = 4 x 6 / 45 = 24/45 = 8/15. Equivalently, red then blue (4/10 x 6/9) plus blue then red (6/10 x 4/9) gives 48/90 = 8/15. The value 4/15 counts only the red-then-blue order. The value 12/25 is the binomial answer 2(0.4)(0.6), which assumes the first chip is replaced. The value 2/5 is the chance that a single chip is red.
Category: Hypergeometric Distribution
Question 17
The number of typos per page in a long report follows a Poisson distribution with an average of 9 typos per page. What is the standard deviation of the number of typos on one page?
3
✓
9
81
4.5
Correct Answer
3
For a Poisson distribution the variance equals the mean, so the variance is 9 and the standard deviation is sqrt(9) = 3. The value 9 is the variance, not its square root. The value 81 squares the mean. The value 4.5 is half the mean, which has no meaning here.
Category: Poisson Distribution
Question 18
A help desk receives an average of 2 calls per hour, and calls follow a Poisson distribution. Using e^-2 = 0.1353, what is the probability of exactly 1 call in a given hour, rounded to two decimal places?
0.14
0.27
✓
0.54
0.73
Correct Answer
0.27
The Poisson formula gives P(X = 1) = (2^1 x e^-2) / 1! = 2 x 0.1353 = 0.2706, which rounds to 0.27. The value 0.14 is e^-2 itself, the probability of zero calls. The value 0.54 doubles the correct answer. The value 0.73 is 1 - 0.27, the probability of any number of calls other than exactly 1.
Category: Poisson Distribution
Question 19
When does the Poisson distribution give a good approximation to a binomial distribution?
When n is small and p is near 0.5
When n is small and p is near 1
When n is large and p is small
✓
When p equals exactly 0.5
Correct Answer
When n is large and p is small
With many trials and a small chance of success on each, successes are rare events, and a Poisson distribution with mean np gives probabilities close to the binomial ones. When n is small and p is near 0.5, the binomial is easy to compute directly and is not like a Poisson. When n is small and p is near 1, successes are common rather than rare. When p equals exactly 0.5, the binomial is symmetric, while a Poisson with a small mean is skewed.
Category: Poisson Distribution
Question 20
Which of these is a discrete random variable?
The time spent waiting for a bus
The height of a randomly chosen student
The weight of a bag of apples
The number of text messages sent in a day
✓
Correct Answer
The number of text messages sent in a day
A count such as the number of text messages takes separate whole-number values, 0, 1, 2 and so on, so it is discrete. The time spent waiting for a bus can take any value in an interval, so it is continuous. The height of a student is a measurement and is continuous. The weight of a bag of apples is also a continuous measurement.
Category: Probability Distributions