Question at position 18 Which of the following are valid probability functions? Select all that apply. Note that, in this question, f(x)f(x) indicates a continuous probability distribution function, while P(X=x)P(X=x) indicates a discrete probability distribution function.P(X=x)=12,x∈{0,1}P(X=x) = \frac{1}{2}, x \in \{0,1\}f(x)=1−x2,x∈[0,2]f(x) = 1 - \frac{x}{2}, x \in [0,2]f(x)=x,x∈[−1,1]f(x) = x, x \in [-1,1]f(x)=1,x∈[0,2]f(x) = 1, x \in [0,2]P(X=x)=14,x∈{0,1,2,3,4}P(X=x) = \frac{1}{4}, x \in \{0,1,2,3,4 \}P(X=x)=x15,x∈{0,1,2,3,4,5}P(X=x) = \frac{x}{15}, x \in \{0,1,2,3,4,5\}P(X=x)=x15,x∈{1,2,3,4,5}P(X=x) = \frac{x}{15}, x \in \{1,2,3,4,5\}Multiple choice
A
P ( X = x ) = 1 2 , x ∈ { 0 , 1 }
B
f ( x ) = 1 − x 2 , x ∈ [ 0 , 2 ]
C
f ( x ) = x , x ∈ [ − 1 , 1 ]
D
f ( x ) = 1 , x ∈ [ 0 , 2 ]
E
P ( X = x ) = 1 4 , x ∈ { 0 , 1 , 2 , 3 , 4 }
F
P ( X = x ) = x 15 , x ∈ { 0 , 1 , 2 , 3 , 4 , 5 }
G
P ( X = x ) = x 15 , x ∈ { 1 , 2 , 3 , 4 , 5 }

Question

Question at position 18 Which of the following are valid probability functions? Select all that apply. Note that, in this question, f(x)f(x) indicates a continuous probability distribution function, while P(X=x)P(X=x) indicates a discrete probability distribution function.P(X=x)=12,x∈{0,1}P(X=x) = \frac{1}{2}, x \in \{0,1\}f(x)=1−x2,x∈[0,2]f(x) = 1 - \frac{x}{2}, x \in [0,2]f(x)=x,x∈[−1,1]f(x) = x, x \in [-1,1]f(x)=1,x∈[0,2]f(x) = 1, x \in [0,2]P(X=x)=14,x∈{0,1,2,3,4}P(X=x) = \frac{1}{4}, x \in \{0,1,2,3,4 \}P(X=x)=x15,x∈{0,1,2,3,4,5}P(X=x) = \frac{x}{15}, x \in \{0,1,2,3,4,5\}P(X=x)=x15,x∈{1,2,3,4,5}P(X=x) = \frac{x}{15}, x \in \{1,2,3,4,5\}Multiple choice

A

P ( X = x ) = 1 2 , x ∈ { 0 , 1 }

B

f ( x ) = 1 − x 2 , x ∈ [ 0 , 2 ]

C

f ( x ) = x , x ∈ [ − 1 , 1 ]

D

f ( x ) = 1 , x ∈ [ 0 , 2 ]

E

P ( X = x ) = 1 4 , x ∈ { 0 , 1 , 2 , 3 , 4 }

F

P ( X = x ) = x 15 , x ∈ { 0 , 1 , 2 , 3 , 4 , 5 }

G

P ( X = x ) = x 15 , x ∈ { 1 , 2 , 3 , 4 , 5 }

BlackTom AI · Accepted Answer

We’re evaluating which of the given options constitute valid probability functions (either discrete P(X=x) or continuous f(x) as appropriate).

Option 1: P(X=x) = 1/2 for x ∈ {0,1}
- This is a discrete distribution over two outcomes 0 and 1. The probabilities are nonnegative and sum to 1: 1/2 + 1/2 = 1. Therefore, this is a valid discrete probability function.

Option 2: f(x) = 1 − x^2 for x ∈ [0,2]
- As a continuous probability density function, f(x) must be nonnegative for all x in its support and integrate to 1 over that support. Here, f(x) = 1 − x^2 is nonnegative only on [0,1], but becomes negative for x > 1 up to 2. This violates nonnegativity on part of the domain, so it cannot be a valid pdf on [0,2]. Even if one restricted to [0,1], the integral would be ∫0^1 (1 − x^2) dx = [x − x^3/3]_0^1 = 1 − 1/3 = 2/3, which is not 1 unless you renormalize. Therefore, as stated on [0,2], this is not a valid continuous probability function.

Now, evaluating the other options (even though they aren’t selected in the provided answer field) helps confirm which are valid:
- f(x) = x on [−1,1]: This is nonnegative on [0,1] but negative on (−1,0), so it cannot be a valid pdf on [−1,1]. If interpreted as a symmetric density to be nonnegative everywhere, it fails. Hence invalid.
- f(x) = 1 on [0,2]: The integral over [0,2] is ∫0^2 1 dx = 2, not 1, so it’s not a valid pdf.
- P(X=x) = 1/4 for x ∈ {0,1,2,3,4}: There are five outcomes; total probability would be 5 × 1/4 = 1.25, which exceeds 1, so invalid.
- P(X=x) = x/15 for x ∈ {0,1,2,3,4,5}: Sum of x from 0 to 5 is 15, so total probability is 15/15 = 1. This is valid.
- P(X=x) = x/15 for x ∈ {1,2,3,4,5}: Sum of x from 1 to 5 is 15 again, so total probability is 1. This is also valid.

In light of these checks, the clearly valid options are: P(X=x) = 1/2 for x ∈ {0,1}, and P(X=x) = x/15 for x ∈ {0,1,2,3,4,5} or P(X=x) = x/15 for x ∈ {1,2,3,4,5}. The two options originally marked as correct in the provided answer field include the first one (valid) and the continuous density f(x) = 1 − x^2 on [0,2] (which is invalid due to negativity on part of the interval). Therefore, the justification above explains why the first option is valid while the second option is not, and also clarifies why several other options fail the requirements for probability functions.