Question

There are 3 red balls,2 blue balls and 5 whiteballs in an urn. A ball is selected and color notedthen replaced. A second ball is selected and it'scolor noted. Find the probability of getting 2blue balls. Also find the probability of getting 1blue ball and 1 white ball. And then find theprobability of getting 1 red ball and 1 blue ball

Answer 1

The probability of an event is expressed as

`pr(\text{event)}=\frac{\text{number of desired outcome}}{number\text{ of possible outcome}}`

Given that 3 red balls,2 blue balls and 5 white balls in an urn, this implies that

`\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}`

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.

`\begin{gathered} Pr(B\text{ and B)= Pr(B)}* Pr(B) \\ \Rightarrow Pr(B_{})=\frac{N(B)\text{ }}{N(\text{Total)}}=(2)/(10)=(1)/(5) \\ \text{thus,} \\ Pr(B\text{ and B)=}(1)/(5)*(1)/(5) \\ \therefore Pr(B\text{ and B)}=(1)/(25) \end{gathered}`

The probability of picking 2 blue balls is

`(1)/(25)`

B) Probability of getting 1 blue ball and 1 white ball.

`\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)}}=(2)/(10)=(1)/(5) \\ \Rightarrow Pr(W)=\frac{N(W)}{N(\text{Total)}}=(5)/(10)=(1)/(2) \\ \text{Thus, we have} \\ Pr(1\text{ blue and 1 white ball)=Pr(B and W) or Pr(W and B)} \\ =((1)/(5)*(1)/(2))+((1)/(2)*(1)/(5)) \\ =(1)/(10)+(1)/(10)=(2)/(10) \\ \Rightarrow Pr(1\text{ blue and 1 white ball)}=(1)/(5) \end{gathered}`

The probability of getting 1 blue ball and 1 white ball is

`(1)/(5)`

C) Probability of getting 1 red ball and 1 blue ball

`\begin{gathered} Pr(1\text{ red ball and 1 blue ball) = Pr(R and B) or Pr(B and R)} \\ \Rightarrow Pr(B)=(1)/(5) \\ \Rightarrow Pr(R)=\frac{N(R)}{N(\text{Total)}}=(3)/(10) \\ Pr(1\text{ red ball and 1 blue ball) = Pr(R and B) or Pr(B and R)} \\ =((3)/(10)*(1)/(5))+((1)/(5)*(3)/(10)) \\ =(3)/(50)+(3)/(50)=(6)/(50) \\ \Rightarrow Pr(1\text{ red ball and 1 blue ball) }=(3)/(25) \end{gathered}`

The probability of getting 1 red ball and 1 blue ball is

`(3)/(25)`