Question

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

Answer 1

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 $\frac{4}{14}$

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