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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

Respuesta :

nevergonnahelp

Answer:

1.79% chance or [tex]\frac{6}{343}[/tex]

Step-by-step explanation:

7+3+4=14

chance of yellow [tex]\frac{3}{14}[/tex]

chance of orange [tex]\frac{4}{14}[/tex]

[tex]\frac{3}{14} * \frac{4}{14} *\frac{4}{14} = \frac{6}{343}[/tex]

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