Suppose that the inside bottom of a box is painted with three colors: 1/3 of the bottom area is red, 1/6 is blue, and 1/2 is white. You toss a tiny pebble into the box without aiming and note the color on which the pebble lands. Then you toss another tiny pebble into the box without aiming and note the color on which that pebble lands. What is the probability that one of the pebbles lands on the color blue and the other pebble lands on red? (Enter your probability as a fraction. Hint: How is this exercise different from finding the probability that the first pebble lands on the color blue and the second pebble lands on red?)

Using the concept of probability, it is found that there is a [tex]\mathbf{\frac{1}{9}}[/tex] probability that one of the pebbles lands on the color blue and the other pebble lands on red.

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A probability is the number of desired outcomes divided by the number of total outcomes.

The probability of the first one landing in blue is 1/6, as blue is 1/6 of the area.
The probability of the second one landing in red is 1/3, as it is 1/3 of the area.
Red then blue also works, with probabilities of 1/3 then 1/6, thus, the probability is multiplied by 2.

[tex]p = 2 \times \frac{1}{6} \times \frac{1}{3} = \frac{2}{6 \times 3} = \frac{1}{3 \times 3} = \frac{1}{9}[/tex]

[tex]\mathbf{\frac{1}{9}}[/tex] probability that one of the pebbles lands on the color blue and the other pebble lands on red.

A similar problem is given at https://brainly.com/question/13831961