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Sagot :
To determine the probability distribution [tex]\( P_X(x) \)[/tex] for the number of times blue occurs when a spinner divided into two equal parts, one red (R) and one blue (B), is spun twice, let's follow a systematic approach.
### Step-by-Step Solution
1. Identify Possible Outcomes:
The possible outcomes when the spinner is spun twice are:
[tex]\[ S = \{ RR, RB, BR, BB \} \][/tex]
2. Define the Random Variable [tex]\(X\)[/tex]:
Let [tex]\( X \)[/tex] be the random variable representing the number of times blue (B) occurs in the outcome.
3. Count the Occurrences of Blue:
- For outcome [tex]\( RR \)[/tex]: Blue appears [tex]\( 0 \)[/tex] times.
- For outcome [tex]\( RB \)[/tex]: Blue appears [tex]\( 1 \)[/tex] time.
- For outcome [tex]\( BR \)[/tex]: Blue appears [tex]\( 1 \)[/tex] time.
- For outcome [tex]\( BB \)[/tex]: Blue appears [tex]\( 2 \)[/tex] times.
4. Determine [tex]\(P(X = x)\)[/tex]:
Each outcome has the same probability of occurring since the regions are equally divided. Thus, the probability of each outcome is:
[tex]\[ P(\text{each outcome}) = \frac{1}{4} \][/tex]
- [tex]\( P(X = 0) \)[/tex]: The event [tex]\( X = 0 \)[/tex] occurs when the outcome is [tex]\( RR \)[/tex].
[tex]\[ P(X = 0) = P(RR) = \frac{1}{4} \][/tex]
- [tex]\( P(X = 1) \)[/tex]: The event [tex]\( X = 1 \)[/tex] occurs when the outcome is [tex]\( RB \)[/tex] or [tex]\( BR \)[/tex].
[tex]\[ P(X = 1) = P(RB) + P(BR) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \][/tex]
- [tex]\( P(X = 2) \)[/tex]: The event [tex]\( X = 2 \)[/tex] occurs when the outcome is [tex]\( BB \)[/tex].
[tex]\[ P(X = 2) = P(BB) = \frac{1}{4} \][/tex]
5. Construct the Probability Distribution Table:
Given these probabilities, the probability distribution [tex]\( P_X(x) \)[/tex] is:
[tex]\[ \begin{array}{|c|c|} \hline X & P_X(x) \\ \hline 0 & 0.25 \\ \hline 1 & 0.5 \\ \hline 2 & 0.25 \\ \hline \end{array} \][/tex]
### Conclusion
Among the given options, the table that matches the probability distribution we calculated is:
[tex]\[ \begin{array}{|c|c|} \hline X & P_X(x) \\ \hline 0 & 0.25 \\ \hline 1 & 0.5 \\ \hline 2 & 0.25 \\ \hline \end{array} \][/tex]
Therefore, the correct probability distribution is the first table in the provided options.
### Step-by-Step Solution
1. Identify Possible Outcomes:
The possible outcomes when the spinner is spun twice are:
[tex]\[ S = \{ RR, RB, BR, BB \} \][/tex]
2. Define the Random Variable [tex]\(X\)[/tex]:
Let [tex]\( X \)[/tex] be the random variable representing the number of times blue (B) occurs in the outcome.
3. Count the Occurrences of Blue:
- For outcome [tex]\( RR \)[/tex]: Blue appears [tex]\( 0 \)[/tex] times.
- For outcome [tex]\( RB \)[/tex]: Blue appears [tex]\( 1 \)[/tex] time.
- For outcome [tex]\( BR \)[/tex]: Blue appears [tex]\( 1 \)[/tex] time.
- For outcome [tex]\( BB \)[/tex]: Blue appears [tex]\( 2 \)[/tex] times.
4. Determine [tex]\(P(X = x)\)[/tex]:
Each outcome has the same probability of occurring since the regions are equally divided. Thus, the probability of each outcome is:
[tex]\[ P(\text{each outcome}) = \frac{1}{4} \][/tex]
- [tex]\( P(X = 0) \)[/tex]: The event [tex]\( X = 0 \)[/tex] occurs when the outcome is [tex]\( RR \)[/tex].
[tex]\[ P(X = 0) = P(RR) = \frac{1}{4} \][/tex]
- [tex]\( P(X = 1) \)[/tex]: The event [tex]\( X = 1 \)[/tex] occurs when the outcome is [tex]\( RB \)[/tex] or [tex]\( BR \)[/tex].
[tex]\[ P(X = 1) = P(RB) + P(BR) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \][/tex]
- [tex]\( P(X = 2) \)[/tex]: The event [tex]\( X = 2 \)[/tex] occurs when the outcome is [tex]\( BB \)[/tex].
[tex]\[ P(X = 2) = P(BB) = \frac{1}{4} \][/tex]
5. Construct the Probability Distribution Table:
Given these probabilities, the probability distribution [tex]\( P_X(x) \)[/tex] is:
[tex]\[ \begin{array}{|c|c|} \hline X & P_X(x) \\ \hline 0 & 0.25 \\ \hline 1 & 0.5 \\ \hline 2 & 0.25 \\ \hline \end{array} \][/tex]
### Conclusion
Among the given options, the table that matches the probability distribution we calculated is:
[tex]\[ \begin{array}{|c|c|} \hline X & P_X(x) \\ \hline 0 & 0.25 \\ \hline 1 & 0.5 \\ \hline 2 & 0.25 \\ \hline \end{array} \][/tex]
Therefore, the correct probability distribution is the first table in the provided options.
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