Questions: Caldwell Community Co Students Home CCCTI-2024FA-MAT-15 Test Ch 3 an myopenmath.com/assess2/?cid=248164aid=17416417#/skip/8 Test Ch 3 and 4 Score: 18.6/63 Answered: 7/22 Question 8 The table contains the number of MM's of each color that were found in a case. MM Distribution Blue Brown Green Orange Red Yellow Total 375 434 547 507 453 369 2685 Find the probability of choosing each color based on this experiment. Round to four decimal places, if necessary. P( blue )= P( brown )= P( green )= P(orange) = P( red )= P( yellow )=

Caldwell Community Co
Students Home
CCCTI-2024FA-MAT-15
Test Ch 3 an
myopenmath.com/assess2/?cid=248164aid=17416417#/skip/8

Test Ch 3 and 4
Score: 18.6/63 Answered: 7/22

Question 8

The table contains the number of MM's of each color that were found in a case.
MM Distribution
Blue Brown Green Orange Red Yellow Total
375 434 547 507 453 369 2685

Find the probability of choosing each color based on this experiment. Round to four decimal places, if necessary.
P( blue )=
P( brown )=
P( green )=
P(orange) =
P( red )=
P( yellow )=

Transcript text: Caldwell Community Co Students Home CCCTI-2024FA-MAT-15 Test Ch 3 an myopenmath.com/assess2/?cid=248164\&aid=17416417\#/skip/8 Test Ch 3 and 4 Score: 18.6/63 Answered: 7/22 Question 8 The table contains the number of M\&M's of each color that were found in a case. M\&M Distribution \begin{tabular}{|c|c|c|c|c|c|c|} \hline Blue & Brown & Green & Orange & Red & Yellow & Total \\ \hline 375 & 434 & 547 & 507 & 453 & 369 & 2685 \\ \hline \end{tabular} Find the probability of choosing each color based on this experiment. Round to four decimal places, if necessary. $P($ blue $)=$ $\square$ $P($ brown $)=$ $\square$ $P($ green $)=$ $\square$ $P$ (orange) $=$ $\square$ $P($ red $)=$ $\square$ $P($ yellow $)=$ $\square$ Download CSV Check Answer

Solution

Solution Steps

To find the probability of choosing each color of M&M, divide the number of M&Ms of each color by the total number of M&Ms. This will give the probability for each color. Ensure the results are rounded to four decimal places.

Step 1: Calculate Total M&Ms

The total number of M&Ms is given as $ 2685 $.

Step 2: Calculate Probability for Each Color

The probability $ P $ of choosing each color is calculated using the formula:

\[ P(\text{color}) = \frac{\text{Number of M&Ms of that color}}{\text{Total number of M&Ms}} \]

Calculating for each color:

For blue: \[ P(\text{blue}) = \frac{375}{2685} \approx 0.1397 \]
For brown: \[ P(\text{brown}) = \frac{434}{2685} \approx 0.1616 \]
For green: \[ P(\text{green}) = \frac{547}{2685} \approx 0.2037 \]
For orange: \[ P(\text{orange}) = \frac{507}{2685} \approx 0.1888 \]
For red: \[ P(\text{red}) = \frac{453}{2685} \approx 0.1687 \]
For yellow: \[ P(\text{yellow}) = \frac{369}{2685} \approx 0.1374 \]

Final Answer

The probabilities for each color are as follows:

$ P(\text{blue}) = 0.1397 $
$ P(\text{brown}) = 0.1616 $
$ P(\text{green}) = 0.2037 $
$ P(\text{orange}) = 0.1888 $
$ P(\text{red}) = 0.1687 $
$ P(\text{yellow}) = 0.1374 $

Thus, the final boxed answers are: \[ \boxed{P(\text{blue}) = 0.1397} \] \[ \boxed{P(\text{brown}) = 0.1616} \] \[ \boxed{P(\text{green}) = 0.2037} \] \[ \boxed{P(\text{orange}) = 0.1888} \] \[ \boxed{P(\text{red}) = 0.1687} \] \[ \boxed{P(\text{yellow}) = 0.1374} \]

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