A bag contains 4 blue marbles, 4 red marbles and 4 green marbles. What's the probability of drawing 2 green marbles WITH replacement? ** Remember P(A and B) = P(A) x P(B)

Question 5 options:

a.1/9

b.1/6

c.1/4

d.1/2

If we look at the problem given, we notice there are 12 marbles in total and 4 green. We want to find the probability of drawing green with replacement.

With that said, the work would be as shown:

[tex]P(A and B) = P(A) * P(B)\\P(A) = \frac{4}{12} \\P(B) = \frac{4}{12} \\\frac{4}{12} * \frac{4}{12} = \frac{16}{144} \\ \frac{16}{144} / 16 = \frac{1}{9}[/tex]

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A bag contains 4 blue marbles, 4 red marbles and 4 green marbles. What's the probability of drawing 2 green marbles WITH replacement? ** Remember P(A and B) = P(A) x P(B) Question 5 options: a.1/9 b.1/6 c.1/4 d.1/2

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A bag contains 4 blue marbles, 4 red marbles and 4 green marbles. What's the probability of drawing 2 green marbles WITH replacement? ** Remember P(A and B) = P(A) x P(B)

Question 5 options:

a.1/9

b.1/6

c.1/4

d.1/2