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 heads in each outcome. for example, if the outcome is hhh, then n(hhh) = 3.

suppose that the random variable x is defined in terms of n as follows: \\(x = 4n - 2n^2 - 1\\). the values of x are given in the table below.

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:

Identify the sample space and outcomes

Using the Sample Space knowledge point
The sample space \(S\) for tossing a fair coin 3 times consists of \(2^3 = 8\) equally likely outcomes:

$$ S = \{tht, hth, hhh, htt, ttt, tth, hht, thh\} $$

Each outcome has a probability of \(\frac{1}{8}\).

Determine the unique values of X

Using the Sample Space knowledge point
From the provided table, we extract the value of \(X\) for each outcome:

• \(X(tht) = 1\)
• \(X(hth) = -1\)
• \(X(hhh) = -7\)
• \(X(htt) = 1\)
• \(X(ttt) = -1\)
• \(X(tth) = 1\)
• \(X(hht) = -1\)
• \(X(thh) = -1\)

The unique values of \(X\), ordered from smallest to largest, are:

$$ x_1 = -7, \quad x_2 = -1, \quad x_3 = 1 $$

Calculate the probability for each value of X

Using the Probability of Events knowledge point
We count the occurrences of each unique value of \(X\) among the 8 outcomes to find \(P(X = x)\):

• For \(x = -7\): Only \(hhh\) gives \(X = -7\).

$$ P(X = -7) = \frac{1}{8} $$

• For \(x = -1\): The outcomes are \(hth, ttt, hht, thh\) (4 outcomes).

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

• For \(x = 1\): The outcomes are \(tht, htt, tth\) (3 outcomes).

$$ P(X = 1) = \frac{3}{8} $$

Construct the probability distribution table

Using the Probability of Events knowledge point
We arrange the values of \(X\) in ascending order along with their corresponding probabilities:

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Answer:

The completed probability distribution table for \(X\) is:

Value \(x\) of \(X\)	\(-7\)	\(-1\)	\(1\)