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2icy_m
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Tyro
Asked: 2021-04-11T11:18:26+00:00 In: ,

Bag containing red, green and blue balls

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A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

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  2. acedstud
    Best Answer

    The probability that none of the balls drawn is blue can be approached in many ways. I will show you two ways you can solve this problem.

    Method 1

    Let X be a random variable representing “picking a blue ball” in a random pick from 2 red, 3 green, and 2 blue balls. Then, using binomial distribution, p=\dfrac{2}{2+3+2}=\dfrac{2}{7}. The probability that X is equal to a value x is given by P(X=x)=^nC_xp^x(1-p)^{n-x}.

    Here, there are two draws of a ball, so n=2. Thus, the probability that none of the balls drawn is blue (x=0), is given by P(X=0)=^2C_0\left(\dfrac{2}{7}\right)^0\left(1-\dfrac{2}{7}\right)^{2-0}

    =1\times1\times\left(\dfrac{5}{7}\right)^2=\dfrac{25}{49}\approx0.5102.

     

    Method 2

    The probability that ‘the first’ ball is not blue is 1 minus the probability of blue = 1-\dfrac{2}{7}=\dfrac{5}{7}.

    The probability that ‘the second’ ball is not blue is 1 minus the probability of blue = 1-\dfrac{2}{7}=\dfrac{5}{7}.

    The probability that ‘the first’ ball is not blue AND ‘the second’ ball is not blue  = \dfrac{5}{7}\times\dfrac{5}{7}=\dfrac{25}{49}\approx0.5102.

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