A bag contains 3 red balls and 1 blue ball. A second bag contains 1 red ball and 1 blue ball. A ball is randomly picked from each bag and is

Question

A bag contains 3 red balls and 1 blue ball. A second bag contains 1 red ball and 1 blue ball. A ball is randomly picked from each bag and is then placed in the other bag. What is the expected number of red balls in the first bag? (mean or expected value)

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Thu Nguyệt 5 years 2021-08-29T03:35:29+00:00 1 Answers 59 views 0

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  1. thucuc
    0
    2021-08-29T03:36:42+00:00
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    Answer:

    “2.82” seems to be the correct solution.

    Step-by-step explanation:

    As we know,

    A bag (Bag A) contains,

    Red balls = 3

    Blue ball = 1

    A second bag (Bag B) contains,

    Red balls = 1

    Blue ball = 1

    Now,

    ⇒ P(Bag A, red ball) = P(A)\times P(r)\times P(r)

    On substituting the values, we get

    ⇒                                = \frac{1}{2}\times \frac{3}{4}\times \frac{2}{3}

    ⇒                                = \frac{1}{4}

    Similarly,

    Probability of second bag will be:

    ⇒  P(Bag B) = \frac{1}{2}

    hence,

    In the first bag (Bag A), the expected number of red balls will be:

    ⇒  E(n)=\Sigma P(n)\times n

    On substituting the values, we get

    ⇒           =3 (\frac{1}{4})+2(\frac{1}{8} )+4(\frac{1}{24} )+3(\frac{1}{12} )+3(\frac{4}{20} )+4(\frac{1}{20} )+2(\frac{3}{20} )+\frac{2}{20}

    ⇒           = \frac{169}{60}

    ⇒           =2.82

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