Respuesta :
Solution:
The probability of an event is expressed as
[tex]pr(\text{event)}=\frac{\text{number of desired outcome}}{number\text{ of possible outcome}}[/tex]Given that 3 red balls,2 blue balls and 5 white balls in an urn, this implies that
[tex]\begin{gathered} \text{Number of red }\Rightarrow\text{N(R)=}3 \\ Number\text{ of blue balls}\Rightarrow N(B)=2 \\ \text{Number of white}\Rightarrow N(W)=5 \\ \text{Total number of balls}\Rightarrow N(Total)=10 \end{gathered}[/tex]A ball is selected and color noted then replaced, a second ball is selected and its color noted.
A) Probability of getting 2 blue balls.
[tex]\begin{gathered} Pr(B\text{ and B)= Pr(B)}\times Pr(B) \\ \Rightarrow Pr(B_{})=\frac{N(B)\text{ }}{N(\text{Total)}}=\frac{2}{10}=\frac{1}{5} \\ \text{thus,} \\ Pr(B\text{ and B)=}\frac{1}{5}\times\frac{1}{5} \\ \therefore Pr(B\text{ and B)}=\frac{1}{25} \end{gathered}[/tex]The probability of picking 2 blue balls is
[tex]\frac{1}{25}[/tex]B) Probability of getting 1 blue ball and 1 white ball.
[tex]\begin{gathered} Pr(1\text{ blue and 1 white ball)=Pr(B and W) or Pr(W and B)} \\ \Rightarrow Pr(B)=\frac{N(B)}{N(\text{Total)}}=\frac{2}{10}=\frac{1}{5} \\ \Rightarrow Pr(W)=\frac{N(W)}{N(\text{Total)}}=\frac{5}{10}=\frac{1}{2} \\ \text{Thus, we have} \\ Pr(1\text{ blue and 1 white ball)=Pr(B and W) or Pr(W and B)} \\ =(\frac{1}{5}\times\frac{1}{2})+(\frac{1}{2}\times\frac{1}{5}) \\ =\frac{1}{10}+\frac{1}{10}=\frac{2}{10} \\ \Rightarrow Pr(1\text{ blue and 1 white ball)}=\frac{1}{5} \end{gathered}[/tex]The probability of getting 1 blue ball and 1 white ball is
[tex]\frac{1}{5}[/tex]C) Probability of getting 1 red ball and 1 blue ball
[tex]\begin{gathered} Pr(1\text{ red ball and 1 blue ball) = Pr(R and B) or Pr(B and R)} \\ \Rightarrow Pr(B)=\frac{1}{5} \\ \Rightarrow Pr(R)=\frac{N(R)}{N(\text{Total)}}=\frac{3}{10} \\ Pr(1\text{ red ball and 1 blue ball) = Pr(R and B) or Pr(B and R)} \\ =(\frac{3}{10}\times\frac{1}{5})+(\frac{1}{5}\times\frac{3}{10}) \\ =\frac{3}{50}+\frac{3}{50}=\frac{6}{50} \\ \Rightarrow Pr(1\text{ red ball and 1 blue ball) }=\frac{3}{25} \end{gathered}[/tex]The probability of getting 1 red ball and 1 blue ball is
[tex]\frac{3}{25}[/tex]