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Answered

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Example 1.29 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 there is a 0.1541 = 15.41% probability that the first three balls are drawn out of the box alternated in colour.

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 with the probabilities given as follows:

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

Hence the probability that the first three balls are drawn out of the box alternated in colour is given by:

[tex]p = \frac{5}{8} \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].

0.1541 = 15.41% probability that the first three balls are drawn out of the box alternated in colour.

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

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