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An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t), which we write [tex]$hth$[/tex], [tex]$ttt$[/tex], etc.

For each outcome, let [tex]$N$[/tex] be the random variable counting the number of heads in each outcome. For example, if the outcome is [tex]$hht$[/tex], then [tex]$N(hht)=2$[/tex]. Suppose that the random variable [tex]$X$[/tex] is defined in terms of [tex]$N$[/tex] as follows:

[tex]\[X = 6N - 2N^2 - 4\][/tex]

The values of [tex]$X$[/tex] are given in the table below.

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
Outcome & hht & htt & hth & tth & thh & ttt & tht & hhh \\
\hline
Value of $X$ & 0 & 0 & 0 & 0 & 0 & -4 & 0 & -4 \\
\hline
\end{tabular}
\][/tex]

Calculate the probabilities [tex]$P(X=x)$[/tex] of the probability distribution of [tex]$X$[/tex]. First, fill in the first row with the values of [tex]$X$[/tex]. Then fill in the appropriate probabilities in the second row.

Sagot :

GinnyAnsweridn
To solve this problem, we need to calculate the probabilities [tex]\( P(X=x) \)[/tex] for the given outcomes of the coin tosses.

Given the coin is tossed 3 times, there are [tex]\( 2^3 = 8 \)[/tex] possible outcomes for the sequence of heads ([tex]\(h\)[/tex]) and tails ([tex]\(t\)[/tex]). The outcomes, along with their respective counts of heads ([tex]\(N\)[/tex]) and values of [tex]\(X\)[/tex], are as follows:

| Outcome | [tex]\(N\)[/tex] (Number of Heads) | [tex]\(X = 6N - 2N^2 - 4\)[/tex] |
|---------|-------------------------|-----------------------|
| h h t | 2 | [tex]\(6(2) - 2(2)^2 - 4 = 12 - 8 - 4 = 0\)[/tex] |
| h t t | 1 | [tex]\(6(1) - 2(1)^2 - 4 = 6 - 2 - 4 = 0\)[/tex] |
| h t h | 2 | [tex]\(6(2) - 2(2)^2 - 4 = 12 - 8 - 4 = 0\)[/tex] |
| t t h | 1 | [tex]\(6(1) - 2(1)^2 - 4 = 6 - 2 - 4 = 0\)[/tex] |
| t h h | 2 | [tex]\(6(2) - 2(2)^2 - 4 = 12 - 8 - 4 = 0\)[/tex] |
| t t t | 0 | [tex]\(6(0) - 2(0)^2 - 4 = 0 - 0 - 4 = -4\)[/tex] |
| t h t | 1 | [tex]\(6(1) - 2(1)^2 - 4 = 6 - 2 - 4 = 0\)[/tex] |
| h h h | 3 | [tex]\(6(3) - 2(3)^2 - 4 = 18 - 18 - 4 = -4\)[/tex] |

Now, we organize the values in the table we need to fill:

[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Outcome} & hht & htt & hth & tth & thh & ttt & tht & hhh \\ \hline \text{Value of } X & 0 & 0 & 0 & 0 & 0 & -4 & 0 & -4 \\ \hline \end{array} \][/tex]

Next, we calculate the probabilities [tex]\( P(X=x) \)[/tex]:

1. Probability [tex]\( P(X=0) \)[/tex]:

We have five outcomes where [tex]\( X = 0 \)[/tex]: \{hht, htt, hth, tth, thh, tht\}.

The total number of outcomes is 8.

Thus, [tex]\( P(X=0) = \frac{6}{8} = 0.75 \)[/tex].

2. Probability [tex]\( P(X=-4) \)[/tex]:

We have two outcomes where [tex]\( X = -4 \)[/tex]: \{ttt, hhh\}.

Thus, [tex]\( P(X=-4) = \frac{2}{8} = 0.25 \)[/tex].

Hence, we can fill in the probability distribution for [tex]\(X\)[/tex]:

[tex]\[ \begin{array}{|c|c|c|} \hline X & \text{Probability} \, P(X=x) \\ \hline 0 & 0.75 \\ -4 & 0.25 \\ \hline \end{array} \][/tex]

So we have:

- [tex]\( P(X = 0) = 0.75 \)[/tex]
- [tex]\( P(X = -4) = 0.25 \)[/tex]

This completes the solution.
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