A bowl contains 10 balls, of which 4 are red and 6 are white. Balls are randomly selected with replacement from the bowl until 4 red balls h

Question

A bowl contains 10 balls, of which 4 are red and 6 are white. Balls are randomly selected with replacement from the bowl until 4 red balls have been selected. Let X be the number of white balls drawn before the fourth red ball is selected. Calculate the probability that X is 6.

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Elliana 2 weeks 2021-09-13T05:15:05+00:00 1 Answer 0

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    2021-09-13T05:16:50+00:00

    Answer:

    The probability that X is 6 = 0.2508

    Step-by-step explanation:

    Probability of picking a red ball = 4 / 10 = 0.4

    Probability of picking a white ball = 6 / 10 = 0.6

    Let’s first state that the last ball selected in this scenario will be red. This implies that we selected 6 white balls and 3 red balls in any manner before we selected the last red ball. So we need to find the probability of taking out 6 white balls and 3 red balls, and use permutation to account for the order they get drawn in:

    Total Probability = Probability of 6 white balls * Probability of 3 red balls * number of ways this can happen

    Probability of 6 white balls = 0.6 ^ 6

    Probability of 3 red balls = 0.4 ^ 3

    Number of ways this can happen (permutation with similar objects) = \frac{9!}{6!*3!}

    Number of ways this can happen = 84

    Total probability = 0.6^6 * 0.4^3 * 84

    Total probability = 0.2508

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45:7+7-4:2-5:5*4+35:2 =? ( )