Suppose 4 blue and 4 red chips are in a hat. Each time we draw a chip we look at its color. If it is blue, we replace it along with one new

Question

Suppose 4 blue and 4 red chips are in a hat. Each time we draw a chip we look at its color. If it is blue, we replace it along with one new blue chip. If it is red, we replace it along with two new red chips. What is the probability that, in successive drawing of chips, the second one is blue?

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Hailey 2 weeks 2021-11-19T12:35:45+00:00 1 Answer 0 views 0

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    2021-11-19T12:37:25+00:00

    Answer:

    \displaystyle P(A)=\frac{43}{90}\approx 0.48

    Step-by-step explanation:

    Probability

    There are 4 blue chips (B) and 4 red chips (R) in a hat. The event to analyze consists in two drawings from the hat, and it can be done in 4 different ways, as shown in the sample space below:

    \Omega=\{BB,BR,RB,RR\}

    We are required to find the probability that the second chip is blue. There are two favourable results to form the event as follows

    A=\{BB,RB\}

    Let’s analyze the first option BB. There are 4 blue chips out of 8 in total, thus the probability to draw a blue chip is

    \displaystyle \frac{4}{8}=\frac{1}{2}

    Once extracted a blue chip, we replace it and add another blue chip, thus the new contents of the hat is: 4 red chips and 5 blue chips. Now we find the probability to draw a second blue chip (5 out of 9):

    \displaystyle \frac{5}{9}

    The combined probability of both events is

    \displaystyle P(BB)=\frac{1}{2}\cdot \frac{5}{9}=\frac{5}{18}

    Now for the second option RB. The probability to first draw a red chip is

    \displaystyle \frac{4}{8}=\frac{1}{2}

    Once we picked a red chip, we replace it along with two new red chips, so the new contents of the hat is: 4 blue chips and 6 red chips. The probability to draw a blue chip in these conditions is

    \displaystyle \frac{4}{10}=\frac{2}{5}

    The combined probability of both events is

    \displaystyle P(RB)=\frac{1}{2}\cdot \frac{2}{5}=\frac{1}{5}

    The required probability is the sum of both separate options

    \displaystyle P(A)=\frac{5}{18}+\frac{1}{5}=\frac{43}{90}

    \boxed{\displaystyle P(A)=\frac{43}{90}\approx 0.48}

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