Answer to Question #303836 in Statistics and Probability for Tin

Question

A box contains 58 red marbles, 14 yellow marbles, and 28 blue marbles. All the marbles are of the same size. Three marbles are drawn at random from the box, first one, then a second, and then a third. Determine the probability of getting one red marble, one yellow marble, and one blue marble with replacement.

Answer 1

$$58+14+28=100$$

$$P(3\ different)=\dfrac{58}{100}(\dfrac{14}{100})(\dfrac{28}{100})$$

$$+\dfrac{58}{100}(\dfrac{28}{100})(\dfrac{14}{100})+\dfrac{14}{100}(\dfrac{58}{100})(\dfrac{28}{100})$$

$$+\dfrac{14}{100}(\dfrac{28}{100})(\dfrac{58}{100})+\dfrac{28}{100}(\dfrac{58}{100})(\dfrac{14}{100})$$

$$+\dfrac{28}{100}(\dfrac{14}{100})(\dfrac{58}{100})=0.14616$$