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Sagot :
[tex] \large \bold\colorbox{gray}{PROBLEM:} [/tex]
Suppose there are two urns, A and B, containing white and yellow balls. Urn A contains 3 white balls and 2 yellow balls, while Urn B contains 2 white balls and 3 yellow balls.
[tex] \bold\colorbox{gray}{Q1.} [/tex] A random ball is transferred from Urn A to Urn B without knowing its color. If a ball is chosen randomly from Urn B after the transfer, what is the probability that it will be a white ball?
[tex] \bold\colorbox{gray}{SOLUTION:} [/tex]
[tex] \begin{array}{l} \textsf{Let the event that a white ball is drawn from urn} \\ B\textsf{ denoted as }W_B. \\ \\ \textsf{Now we have to consider 2 cases.} \\ \\ \bold{Case\ 1:}\textsf{ A white ball is drawn from urn }A. \\ \\ \textsf{The probability of drawing a white ball from urn} \\ A\textsf{ and then a white ball from urn }B: \\ \\ \quad \quad \dfrac{3}{3 + 2} \times \dfrac{2 + 1}{2 + 3 + 1} = \dfrac{3}{10} \\ \\ \\ \bold{Case\ 2:}\textsf{ A yellow ball is drawn from urn }A. \\ \\ \textsf{The probability of drawing a yellow ball from} \\ \textsf{urn }A\textsf{ and then a white ball from urn }B: \\ \\ \quad \quad \dfrac{2}{3 + 2} \times \dfrac{2}{2 + 3 + 1} = \dfrac{2}{15} \\ \\ \textsf{Thus, the overall probability is}: \\ \\ \quad \: \: P(W_B) = \dfrac{3}{10} + \dfrac{2}{15} = \boxed{\dfrac{13}{30}}\:\:\:\textit{Answer} \end{array} [/tex]
[tex] \bold\colorbox{gray}{Q2.}[/tex] Suppose 2 balls are transferred from Urn A to Urn B without knowing its color. If a ball is chosen randomly from Urn B after the transfer, what is the probability that it will be a white ball?
[tex] \bold\colorbox{gray}{SOLUTION:} [/tex]
[tex] \begin{array}{l} \textsf{Let the event that a white ball is drawn from urn} \\ B\textsf{ denoted as }W_B. \\ \\ \textsf{Now we have to consider 3 cases.} \\ \\ \bold{Case\ 1:}\textsf{ 2 white balls are drawn from urn }A. \\ \\ \textsf{The probability of drawing 2 white balls from} \\ \textsf{urn }A\textsf{ and then a white ball from urn }B: \\ \\ \quad \dfrac{3}{3 + 2} \times \dfrac{3 - 1}{3 + 2 - 1} \times \dfrac{2 + 2}{2 + 3 + 2} = \dfrac{6}{35} \\ \\ \\ \bold{Case\ 2:}\textsf{ A white ball and a yellow ball are drawn} \\ \qquad \quad\:\:\: \textsf{from urn }A. \\ \\ \textsf{The probability of drawing a white ball and a} \\ \textsf{yellow ball from urn }A\textsf{ and then a white ball} \\ \textsf{from urn }B: \\ \\ \quad \dfrac{3}{3 + 2} \times \dfrac{2}{3 + 2 - 1} \times \dfrac{2 + 1}{2 + 3 + 2} = \dfrac{9}{70} \\ \\ \bold{Case\ 3:}\textsf{ 2 yellow balls are drawn from urn }A. \\ \\ \textsf{The probability of drawing 2 yellow balls from} \\ \textsf{urn }A\textsf{ and then a white ball from urn }B: \\ \\ \quad \dfrac{2}{3 + 2} \times \dfrac{2 - 1}{3 + 2 - 1} \times \dfrac{2}{2 + 3 + 2} = \dfrac{1}{35} \\ \\ \textsf{Thus, the overall probability is}: \\ \\ \:\:\: P(W_B) = \dfrac{6}{35} + \dfrac{9}{70} + \dfrac{1}{35} = \boxed{\dfrac{23}{70}}\:\:\:\textit{Answer} \end{array} [/tex]
[tex] \mathfrak \colorbox{gray}{\#CarryOnLearning} [/tex]
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