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12345 [234]
2 years ago
13

Consider an urn with 7 black balls, 3 yellow balls, and 4 orange balls. If 3 balls are chosen randomly with replacement, what is

the probability that the first is yellow, the second is orange, and the third is orange
Mathematics
1 answer:
KATRIN_1 [288]2 years ago
5 0

Answer:

1.79% chance or \frac{6}{343}

Step-by-step explanation:

7+3+4=14

chance of yellow \frac{3}{14}

chance of orange \frac{4}{14}

\frac{3}{14} * \frac{4}{14} *\frac{4}{14}  =  \frac{6}{343}

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\boxed{ \sf{  \color{purple}\huge Answer :}}

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\large \tt \purple↪  \: \:  \:  \:  \:5x - x = 10 + 16

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1 year ago
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hope this helped

Step-by-step explanation:

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