MAT prob&stat1 discrete random variables

Question 1

What is a probability distribution?

Answer 1

A probability distribution lists the exhaustive and mutually exclusive outcomes of an experiment and their associated probabilities.

Question 2

What is a discrete random variable?

Answer 2

A discrete random variable maps an experiment to a set of discrete numerical outcomes, such as X={1,2,3,…}.

Question 3

How is the probability P(X=x) defined?

Answer 3

P(X = x) is the probability that the random variable X takes the value x.

Question 4

What are the properties of a probability distribution?

Answer 4

1. The probabilities for all possible outcomes must sum to 1.
2. Each outcome is mutually exclusive.
3. Probabilities indicate how likely each outcome is.

Question 5

How do you represent a (discrete) probability distribution visually?

Answer 5

Probability distributions can be represented using vertical line graphs, where each line corresponds to an outcome and its probability.

Question 6

What is the probability distribution for summing the values of two spins of a six-sided spinner?

find ans

Answer 6

x: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
P(X = x): 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36, 4/36, 3/36, 2/36, 1/36

Question 7

How do you verify if a (discrete) probability distribution is valid?

Answer 7

Ensure that the sum of all probabilities equals 1, i.e., sum P(X = x) = 1

Question 8

How do you find a constant c in a probability distribution?

Answer 8

Use the condition 𝛴P(X = x) = 1, solve for c, and substitute it back into the probabilities.

Question 9

Example: A discrete random variable X has probabilities P(X = r) = r/15 for r = 1, 3, 4, 7. What is the probability distribution?

find ans

Answer 9

r: 1, 3, 4, 7
P(X = r): 1/15, 3/15, 4/15, 7/15

Question 10

Example: A random variable Y has P(Y = 1) = c, P(Y = 2) = 4c, P(Y = 3) = 3c, P(Y = 4) = 4c. Find c.

Answer 10

Using 𝛴 P(Y = y) = 1:
c + 4c + 3c + 4c = 1 implies 12c = 1, so c = 1/12
The probabilities are 1/12, 4/12, 3/12, 4/12.

Question 11

A bag has 4 blue discs and 3 green discs. Two discs are removed without replacement. What is the probability distribution of X, the number of blue discs removed?

fins ans

Answer 11

x: 0, 1, 2
P(X = x): 1/7, 4/7, 2/7

Question 12

What is the expectation (mean) of a discrete random variable X?

Answer 12

E(X) = Σ [x_i * P(X = x_i)]
Multiply each value of x by its probability, then sum these values.

Question 13

Example: If X has values 0, 1, 2, 3 with probabilities 1/12, 1/3, 1/6, 5/12, what is E(X)?

Answer 13

E(X) = 0 * (1/12) + 1 * (1/3) + 2 * (1/6) + 3 * (5/12) = 23/12

Question 14

What is variance (Var(X)) of a discrete random variable?

Answer 14

Var(X) = E(X^2) - [E(X)]^2
Where E(X^2) = Σ [x_i^2 * P(X = x_i)].

Question 15

Example: Find Var(X) for X = {0, 1, 2, 3} with probabilities 1/12, 1/3, 1/6, 5/12.

Answer 15

E(X) = 23/12
E(X^2) = 19/4
Var(X) = 19/4 - (23/12)^2 = 1.08

Question 16

What is the equivalence of two variance formulas?

Answer 16

Both:
Var(X) = E(X^2) - [E(X)]^2
and
Var(X) = E[(X - μ)^2]
produce the same result.

Question 17

Example: What is E(X) when a six-sided spinner is spun once?

Answer 17

E(X) = Σ [r * (1/6)] for r = 1 to 6
E(X) = (1 + 2 + 3 + 4 + 5 + 6) / 6 = 3.5

Question 18

How do you calculate Var(X) for two spinners with values summed?

Answer 18

1. Tabulate P(X = x) using a sample space.
2. Use E(X) = Σ [x * P(X = x)].
3. Find E(X^2) = Σ [x^2 * P(X = x)].
4. Compute Var(X) = E(X^2) - [E(X)]^2.

Question 19

What is the expectation and variance for two spinners with summed results X = {2, 3, 4, …, 8}?

Answer 19

E(X) = 5
E(X^2) = 27.5
Var(X) = 27.5 - 5^2 = 2.5

Question 20

How do you find a constant k in a probability distribution?

Answer 20

Use the condition:
Σ P(X = x) = 1
Solve for k and substitute back into the probabilities.

Question 21

Example: For P(X = r) = k / (r + 1) with r = 0, 1, 2, 3, find k.

Answer 21

k * (1 + 1/2 + 1/3 + 1/4) = 1
k = 5/9