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A statistics teacher wanted to know how often students go to the movies. She surveyed the 25 students in her class to find out how many movies they saw at a movie theater in the past month. Here are the results:

\begin{tabular}{|l|c|c|c|c|}
\hline
\begin{tabular}{l}
Number of \\
Movies
\end{tabular} & 0 & 1 & 2 & \begin{tabular}{c}
3 or \\
more
\end{tabular} \\
\hline
Probability & 0.72 & 0.20 & 0.04 & [tex]$?$[/tex] \\
\hline
\end{tabular}

Let [tex]$M$[/tex] represent the number of movies that a student saw at a movie theater in the past month.

[tex]\[
P(M \geq 3) = \square
\][/tex]

What is the probability that a randomly selected student saw 1 or fewer movies in the past month?

[tex]\[
\square
\][/tex]

Identify the complement of seeing 1 or fewer movies in the past month. What is the probability of this event?

[tex]\[
\square
\][/tex]



Answer :

GinnyAnswer
Sure, let's solve this step-by-step.

1. Finding the probability that a student saw 3 or more movies in the past month:

From the provided table, we know the probabilities for seeing 0, 1, and 2 movies:
[tex]\[ \begin{aligned} & P(M = 0) = 0.72, \\ & P(M = 1) = 0.20, \\ & P(M = 2) = 0.04. \end{aligned} \][/tex]

To find the probability that a student saw 3 or more movies, we need to sum the probabilities of the complementary events (0, 1, and 2 movies) and subtract from 1:

[tex]\[ P(M \geq 3) = 1 - (P(M = 0) + P(M = 1) + P(M = 2)). \][/tex]

Substituting the values, we get:

[tex]\[ P(M \geq 3) = 1 - (0.72 + 0.20 + 0.04) = 1 - 0.96 = 0.04. \][/tex]

So, the probability that a student saw 3 or more movies in the past month is:
[tex]\[ P(M \geq 3) = 0.04. \][/tex]


2. Finding the probability that a randomly selected student saw 1 or fewer movies in the past month:

To find this, we sum the probabilities of seeing 0 or 1 movies:

[tex]\[ P(M \leq 1) = P(M = 0) + P(M = 1). \][/tex]

Substituting the values, we get:

[tex]\[ P(M \leq 1) = 0.72 + 0.20 = 0.92. \][/tex]

So, the probability that a student saw 1 or fewer movies in the past month is:
[tex]\[ P(M \leq 1) = 0.92. \][/tex]

3. Identifying the complement of seeing 1 or fewer movies in the past month and finding the probability of this event:

The complement of the event "seeing 1 or fewer movies" is "seeing more than 1 movie." Therefore, we need to find:
[tex]\[ P(M > 1). \][/tex]

This can be determined by subtracting the probability of seeing 1 or fewer movies from 1:

[tex]\[ P(M > 1) = 1 - P(M \leq 1). \][/tex]

Substituting the value we found earlier:

[tex]\[ P(M > 1) = 1 - 0.92 = 0.08. \][/tex]

So, the probability of seeing more than 1 movie in the past month is:
[tex]\[ P(M > 1) = 0.08. \][/tex]

To summarize:
- The probability that a student saw 3 or more movies in the past month is [tex]\(0.04\)[/tex].
- The probability that a student saw 1 or fewer movies in the past month is [tex]\(0.92\)[/tex].
- The probability of the complement event, seeing more than 1 movie, is [tex]\(0.08\)[/tex].

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