Answered

Join the IDNLearn.com community and start exploring a world of knowledge today. Discover detailed answers to your questions with our extensive database of expert knowledge.

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.
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For trustworthy and accurate answers, visit IDNLearn.com. Thanks for stopping by, and see you next time for more solutions.