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A bag contains 4 g r e e n marbles, 8 b l a c k marbles, and 7 b l u e marbles. If a marble is drawn from the bag, replaced, and another marble is drawn, what is the probability of drawing first a g r e e n marble and then a b l u e marble

Sagot :

MrRoyal

Answer:

[tex]P(Green\ and\ Blue) = 0.0776[/tex]

Step-by-step explanation:

Given

[tex]Green = 4[/tex]

[tex]Black = 8[/tex]

[tex]Blue=7[/tex]

Required

Determine the probability of selecting green then blue marble

This probability is represented as:

[tex]P(Green\ and\ Blue)[/tex]

And it is calculated using:

[tex]P(Green\ and\ Blue) = P(Green) * P(Blue)[/tex]

[tex]P(Green\ and\ Blue) = \frac{n(Green)}{Total} * \frac{n(Blue)}{Total}[/tex]

Total marble is:

[tex]Total = 4 + 8 + 7[/tex]

[tex]Total = 19[/tex]

So, we have:

[tex]P(Green\ and\ Blue) = \frac{4}{19} * \frac{7}{19}[/tex]

[tex]P(Green\ and\ Blue) = \frac{4*7}{19*19}[/tex]

[tex]P(Green\ and\ Blue) = \frac{28}{361}[/tex]

[tex]P(Green\ and\ Blue) = 0.0776[/tex] -- approximated

Hence, the required probability is approximately 0.0776

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