An urn contains seven red marbles, four white marbles, and five blue marbles. If three marbles are drawn in succession (each being replaced before the next is drawn), what is the probability that the first marble is red, the second is white, and the third is blue?

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Answer:

The required probability is 0.034.

Step-by-step explanation:

Consider the provided information.

An urn contains seven red marbles, four white marbles, and five blue marbles.

Red marbles = 7

White marbles = 4

Blue marble = 5

Total number of marbles = 7+4+5 = 16

Three marbles are drawn in succession.

Probability that the first marble is red, the second is white, and the third is blue is:

[tex]P=\frac{7}{16}\times\frac{4}{16}\times\frac{5}{16}\\\\P=\frac{35}{1024}\\P\approx0.034[/tex]

Hence, the required probability is 0.034.

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