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11. lauren has a bag containing 6 red, 8 blue, 5 green, 9 yellow, and 2…

Question

  1. lauren has a bag containing 6 red, 8 blue, 5 green, 9 yellow, and 2 white marbles that are all the same size and shape. what is the probability or randomly choosing a white marble on the first pick, not replacing it, and then randomly choosing a green marble on the second pick?

Explanation:

Step1: Calculate total marbles initially

First, find the total number of marbles. Add the number of each color: \(6 + 8 + 5 + 9 + 2\).
\(6+8 = 14\), \(14 + 5 = 19\), \(19 + 9 = 28\), \(28 + 2 = 30\). So total marbles \(n = 30\).

Step2: Probability of white first

Number of white marbles \(w = 2\). Probability of choosing white first, \(P(W) = \frac{w}{n}=\frac{2}{30}=\frac{1}{15}\).

Step3: Total marbles after first pick

After picking a white marble (not replacing), total marbles become \(30 - 1 = 29\).

Step4: Probability of green second

Number of green marbles \(g = 5\). Probability of choosing green second, \(P(G|W) = \frac{g}{29}\).

Step5: Multiply probabilities (dependent events)

Since the events are dependent (no replacement), the combined probability is \(P(W \cap G)=P(W)\times P(G|W)=\frac{1}{15}\times\frac{5}{29}\).
Simplify: \(\frac{1\times5}{15\times29}=\frac{5}{435}=\frac{1}{87}\).

Answer:

\(\frac{1}{87}\)