Question

an ordinary (fair) coin is tossed 3 times. outcomes are thus triples of \heads\ (h) and \tails\ (t) which we write hth, ttt, etc. for each outcome, let n be the random variable counting the number of tails in each outcome. for example, if the outcome is htt, then n(htt) = 2. suppose that the random variable x is defined in terms of n as follows: ( x = 2n^2 - 6n - 2 ). the values of x are given in the table below.

outcome	hth	thh	tth	htt	hht	hhh	ttt	tht

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.

value x of x	☐	☐

Explanation:

Step1: Identify unique X values

From the table, the unique values of \( X \) are \(-6\) and \(-2\).

Step2: Count outcomes for each X

• For \( X = -6 \): Count the number of outcomes with \( X = -6 \). From the table, the outcomes \( hrh, thh, tth, htr, hht, tht \) (that's 6 outcomes) and check the total number of possible outcomes when tossing a coin 3 times: \( 2^3 = 8 \) outcomes. Wait, the table has 8 outcomes: \( hrh, thh, tth, htr, hht, hhh, ttt, tht \). So for \( X = -6 \): let's count the outcomes with \( X = -6 \): \( hrh, thh, tth, htr, hht, tht \) → 6 outcomes. For \( X = -2 \): \( hhh, ttt \) → 2 outcomes.

Step3: Calculate probabilities

Probability is number of favorable outcomes over total outcomes (8).

• For \( X = -6 \): \( P(X = -6) = \frac{6}{8} = \frac{3}{4} \)
• For \( X = -2 \): \( P(X = -2) = \frac{2}{8} = \frac{1}{4} \)

Answer:

Value \( x \) of \( X \): \(-6\), \(-2\)
\( P(X = x) \): \(\frac{3}{4}\), \(\frac{1}{4}\)