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Audrey and her best friend wanted to paint their apartment. In their storage closet, they found $8$ cans of paint. Audrey remembered that $5$ of the cans contained beige paint, but none of the cans had labels. $\newline$ If Audrey randomly opened $4$ cans, what is the probability that all of them contain beige paint? $\newline$ Write your answer as a decimal rounded to four decimal places. $\newline$ ____ $\newline$

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Find probabilities using combinations and permutations

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Q. Audrey and her best friend wanted to paint their apartment. In their storage closet, they found

8

cans of paint. Audrey remembered that

5

of the cans contained beige paint, but none of the cans had labels.

\newline

If Audrey randomly opened

4

cans, what is the probability that all of them contain beige paint?

\newline

Write your answer as a decimal rounded to four decimal places.

\newline

____

\newline

Total Cans Calculation: Total number of cans: $8$ $\newline$ Cans to be opened: $4$ $\newline$ Calculate the total number of ways to choose $4$ cans out of $8$ . $\newline$ Total outcomes: $_{8}C_{4}$
Calculate $8C4$ : Find the value of $_8C_4$ .
$_8C_4 = \frac{8!}{4!(8-4)!} = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4! \times 4 \times 3 \times 2 \times 1} = 70$
Beige Paint Cans: Number of beige paint cans: $5$ $\newline$ Cans of beige paint to be opened: $4$ $\newline$ Calculate the number of ways to choose $4$ beige cans out of $5$ . $\newline$ Favorable outcomes: $_{5}C_{4}$
Calculate $5C4$ : Find the value of $_5C_4$ .
$_5C_4 = \frac{5!}{4!(5-4)!} = \frac{5!}{4!1!} = \frac{5 \times 4!}{4! \times 1} = 5$
Calculate Probability: Calculate the probability that all opened cans are beige. $\newline$ Probability = Favorable outcomes / Total outcomes $\newline$ = $\frac{5}{70}$ $\newline$ $\approx 0.0714$

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