Question

a number cube (a fair die) is rolled 3 times. for each roll, we are interested in whether the roll comes up odd or even. an outcome is represented by a string of the sort oee (meaning an odd number on the first roll, an even number on the second roll, and an even number on the third roll).
for each outcome, let ( n ) be the random variable counting the number of even rolls in each outcome. for example, if the outcome is oee, then ( n(\text{oee}) = 2 ).
suppose that the random variable ( x ) is defined in terms of ( n ) as follows: ( x = (n - 1)^2 - 2 ). the values of ( x ) are given in the table below.

outcome	ooe	oeo	oee	eoo	eee	eeo	ooo	eoe

calculate the probabilities ( p(x = x) ) of the probability distribution of ( x ). first, fill in the first row with the values of ( x ). then fill in the appropriate probabilities in the second row.

Explanation:

Step1: Identify unique X values

From the table, the values of \( X \) are \( 0 \) and \( -2 \).

Step2: Count outcomes for each X

• For \( X = 0 \): Outcomes are ooe, oeo, ooo, eoo, eoe, eoo? Wait, let's list all 8 outcomes: ooe, oeo, ooo, eoo, eoe, eee? No, the 8 outcomes are: ooe, oeo, ooo, eoo, eoe, eee, eeo, oee? Wait the given outcomes are: ooe, oeo, oee, eoo, eee, eeo, ooo, eoe (8 outcomes). Now check their X values:
• ooe: 0
• oeo: 0
• oee: 0
• eoo: 0
• eee: -2
• eeo: 0
• ooo: -2
• eoe: 0

So count: \( X = 0 \) has 6 outcomes, \( X = -2 \) has 2 outcomes.

Step3: Calculate probabilities

Total outcomes: 8 (since a fair die rolled 3 times, each roll has 2 options (odd/even), so \( 2^3 = 8 \) outcomes).

• \( P(X = 0) = \frac{\text{Number of outcomes with } X=0}{8} = \frac{6}{8} = \frac{3}{4} \)
• \( P(X = -2) = \frac{\text{Number of outcomes with } X=-2}{8} = \frac{2}{8} = \frac{1}{4} \)

Answer:

Value \( x \) of \( X \)	\( -2 \)	\( 0 \)

Wait, let's recheck the outcomes:

Given outcomes: ooe, oeo, oee, eoo, eee, eeo, ooo, eoe (8 outcomes).

Their X values:

• ooe: 0
• oeo: 0
• oee: 0
• eoo: 0
• eee: -2 (from table: eee has X=-2)
• eeo: 0 (table: eeo has X=0)
• ooo: -2 (table: ooo has X=-2)
• eoe: 0 (table: eoe has X=0)

So yes: X=0: ooe, oeo, oee, eoo, eeo, eoe (6 outcomes), X=-2: eee, ooo (2 outcomes). So probabilities:

\( P(X = -2) = \frac{2}{8} = \frac{1}{4} \)

\( P(X = 0) = \frac{6}{8} = \frac{3}{4} \)

So the probability distribution is:

For \( x = -2 \), \( P(X = -2) = \frac{1}{4} \)

For \( x = 0 \), \( P(X = 0) = \frac{3}{4} \)