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Answered

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Pls give me answer
Initially, a box contains three white and five red balls. Balls
are drawn at random from the box. Whenever a red ball is drawn, it is placed
back into the box together with one extra red ball. Whenever a white ball is
drawn, it is placed back into the box together with two extra white balls. What
is the probability that the first three balls are drawn out of the box alternated
in colour?

Sagot :

Using it's concept, it is found that the probability that the first three balls are drawn out of the box alternated in colour is of 0.1541 = 15.41%.

What is a probability?

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

Considering the situation described, the ways that alternate colored balls can be taken, and their probabilities, are given as follows:

  • R(5/8) - W(3/9) - R(5/11).
  • W(3/8) - R(5/10) - W(5/11).

Hence the probability is given by:

[tex]p = \frac{5}{11} \times \frac{3}{9} \times \frac{5}{11} + \frac{3}{8} \times \frac{5}{10} \times \frac{5}{11} = 0.0689 + 0.0852 = 0.1541[/tex]

More can be learned about probabilities at https://brainly.com/question/14398287

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