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10 June, 12:14

A jar contains four red marbles and six green marbles. You randomly select a marble from the jar, with replacement. The random variable represents the number of red marbles. What is the probability of getting exactly two red marbles out of four trials?

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  1. 10 June, 12:22
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    The probability of getting exactly two read marbles is P = 0,003456%

    Step-by-step explanation:

    So, each of the following sequences are the desired results (I will use R for a red marble and G for a green marble).

    S (1) = R-R-G-G

    S (2) = R-G-R-G

    S (3) = R-G-G-R

    S (4) = G-R-R-G

    S (5) = G-R-G-R

    S (6) = G-G-R-R

    In all, considering there are replacement, there can be 10*10*10*10 = 10000 total sequences, so the probability of getting exactly two read marbles is

    P = / frac{P (S (1)) + P (S (2)) + P (S (3)) + P (S (4)) + P (S (5)) + P (S (6)) }{10000}

    where

    P (S (1)) = P (S (2)) = P (S (3)) = P (S (4)) = P (S (5)) = P (S (6)) = (0.4) ^{2} * (0.6) ^{2} = 0.16*0.36 = 0.0576.

    The probability of getting exactly two read marbles is

    P = / frac{6*0.0576}{10000} = 0,003456%
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