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A group of students at a high school took a standardized test. The number of students who passed or failed the exam is broken down by those who did or did not eat breakfast in the following table. Determine whether eating breakfast and passing the test are independent by filling out the blanks in the sentence below, rounding all probabilities to the nearest thousandth

Passed
Failed

Did Eat Breakfast
38
19

Didn't Eat Breakfast
58
29

Answer Attempt 1 out of 2
Since
P( did eat breakfast
)×P( pass
)=◻ and
P( did eat breakfast and pass
)=◻, the two results are so the events are

Score: $1 / 3$ $\newline$ Penalty: none $\newline$ Question $\newline$ Watch Video $\newline$ Show Examples $\newline$ A group of students at a high school took a standardized test. The number of students who passed or failed the exam is broken down by those who did or did not eat breakfast in the following table. Determine whether eating breakfast and passing the test are independent by filling out the blanks in the sentence below, rounding all probabilities to the nearest thousandth $\newline$ \begin{tabular}{|c|c|c|} $\newline$ \hline & Passed & Failed \\ $\newline$ \hline Did Eat Breakfast & $38$ & $19$ \\ $\newline$ \hline Didn't Eat Breakfast & $58$ & $29$ \\ $\newline$ \hline $\newline$ \end{tabular} $\newline$ Answer Attempt $1$ out of $2$ $\newline$ Since $\mathrm{P}($ did eat breakfast $) \times \mathrm{P}($ pass $)=\square$ and $\mathrm{P}($ did eat breakfast and pass $)=\square$ , the two results are so the events are

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A group of students at a high school took a standardized test. The number of students who passed or failed the exam is broken down by those who did or did not eat breakfast in the following table. Determine whether eating breakfast and passing the test are independent by filling out the blanks in the sentence below, rounding all probabilities to the nearest thousandth

\newline

\begin{tabular}{|c|c|c|}

\newline

\hline & Passed & Failed \\

\newline

\hline Did Eat Breakfast &

38

19

\newline

\hline Didn't Eat Breakfast &

58

29

\newline

\hline

\newline

\end{tabular}

\newline

Answer Attempt

1

out of

2

\newline

Since

\mathrm{P}(

did eat breakfast

) \times \mathrm{P}(

pass

)=\square

and

\mathrm{P}(

did eat breakfast and pass

)=\square

, the two results are so the events are

Calculate Probability of Eating Breakfast: To determine whether eating breakfast and passing the test are independent events, we need to calculate the probability of each event and the joint probability of both events occurring together. We will then compare the product of the individual probabilities to the joint probability. $\newline$ First, let's calculate the probability of a student eating breakfast. We add the number of students who ate breakfast and passed with those who ate breakfast and failed to get the total number of students who ate breakfast. $\newline$ Number of students who ate breakfast and passed = $38$ $\newline$ Number of students who ate breakfast and failed = $19$ $\newline$ Total number of students who ate breakfast = $38 + 19 = 57$ $\newline$ Next, we calculate the total number of students who took the test by adding all the numbers in the table. $\newline$ Total number of students = $38$ (passed and ate breakfast) + $19$ (failed and ate breakfast) + $58$ (passed and didn't eat breakfast) + $29$ (failed and didn't eat breakfast) = $38 + 19 + 58 + 29 = 144$ $\newline$ Now we can find the probability of a student eating breakfast. $\newline$ $P(\text{did eat breakfast}) = \frac{\text{Total number of students who ate breakfast}}{\text{Total number of students}} = \frac{57}{144}$ $\newline$ Let's calculate this probability.
Calculate Probability of Passing Test: Calculating $P(\text{did eat breakfast})$ : $P(\text{did eat breakfast}) = \frac{57}{144} \approx 0.396$ (rounded to the nearest thousandth)Now let's calculate the probability of a student passing the test. We add the number of students who passed and ate breakfast with those who passed and didn't eat breakfast to get the total number of students who passed.Number of students who passed and ate breakfast = $38$ Number of students who passed and didn't eat breakfast = $58$ Total number of students who passed = $38 + 58 = 96$ Now we can find the probability of a student passing the test. $P(\text{pass}) = \frac{\text{Total number of students who passed}}{\text{Total number of students}} = \frac{96}{144}$ Let's calculate this probability.
Calculate Joint Probability: Calculating $P(\text{pass})$ : $P(\text{pass}) = \frac{96}{144} \approx 0.667$ (rounded to the nearest thousandth)Next, we need to calculate the joint probability of a student both eating breakfast and passing the test.Number of students who ate breakfast and passed = $38$ Now we can find the joint probability. $P(\text{did eat breakfast and pass}) = \frac{\text{Number of students who ate breakfast and passed}}{\text{Total number of students}} = \frac{38}{144}$ Let's calculate this joint probability.
Check Independence of Events: Calculating $P(\text{did eat breakfast and pass})$ : $P(\text{did eat breakfast and pass}) = \frac{38}{144} \approx 0.264$ (rounded to the nearest thousandth) $\newline$ Now we have all the probabilities we need to determine if the events are independent. If the events are independent, then the following equation should hold true: $P(\text{did eat breakfast}) \times P(\text{pass}) = P(\text{did eat breakfast and pass})$ $\newline$ Let's check if this is the case by multiplying $P(\text{did eat breakfast})$ and $P(\text{pass})$ .
Check Independence of Events: Calculating $P(\text{did eat breakfast and pass})$ : $P(\text{did eat breakfast and pass}) = \frac{38}{144} \approx 0.264$ (rounded to the nearest thousandth)Now we have all the probabilities we need to determine if the events are independent. If the events are independent, then the following equation should hold true: $P(\text{did eat breakfast}) \times P(\text{pass}) = P(\text{did eat breakfast and pass})$ Let's check if this is the case by multiplying $P(\text{did eat breakfast})$ and $P(\text{pass})$ .Multiplying $P(\text{did eat breakfast})$ and $P(\text{pass})$ : $P(\text{did eat breakfast}) \times P(\text{pass}) \approx 0.396 \times 0.667 \approx 0.264$ (rounded to the nearest thousandth)Now we compare this product to the joint probability $P(\text{did eat breakfast and pass})$ .Since the product of $P(\text{did eat breakfast})$ and $P(\text{pass})$ is approximately equal to $P(\text{did eat breakfast and pass})$ , the two events are independent.Therefore, eating breakfast and passing the test are independent events.