Question

12. Refer to a bag containing 13 red balls numbered 1-13 and 5 green balls numbered 14-18. Youchoose a ball at random.aa. What is the probability that you choose a red or even numbered ball? (3 points)b. What is the probability you choose a green ball or a ball numbered less than 5?(3 points)

Answer 1

Given

A bag contains 13 red balls numbered 1-13 and 5 green balls numbered 14-18.

And a ball is chosen at random.

To find:

a) The probability that you choose a ren or even numbered ball.

b) The the probability you choose a green ball or a ball numbered less than 5.

Step-by-step explanation:

It is given that,

The total number of balls is (13+5)=18.

The number of red balls is 13 numbered from 1-13.

The number of green balls is 5 numbered from 14-18.

i) Then, the probability of getting a a red or even numbered ball is,

`\begin{gathered} P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right) \\ =(n\left(A\right))/(n\left(S\right))+(n\left(B\right))/(n\left(S\right))-(n\left(A\cap B\right))/(n\left(S\right)) \end{gathered}`

Here,

`\begin{gathered} n\left(A\right)=n\left(red\text{ balls}\right) \\ =13 \\ n\left(B\right)=n\left(even\text{ numbered balls}\right) \\ =9 \\ n\left(A\cap B\right)=n\left(red\text{ balls and even numbered balls}\right) \\ =6 \\ n\left(S\right)=n\left(Total\text{ number of balls}\right) \\ =18 \end{gathered}`

That implies,

`\begin{gathered} P\left(A\cup B\right)=(13)/(18)+(9)/(18)-(6)/(18) \\ =(16)/(18) \\ =(8)/(9) \\ =0.89 \end{gathered}`

Hence, the probability of getting a red or even numbered ball is 0.89.

ii) Also,

The probability of getting a green ball or a ball numbered less than 5 is,

`\begin{gathered} P\left(C\cup D\right?=P\left(C\right)+P\left(D\right)-P\left(C\cap D\right) \\ =(n\left(C\right))/(n\left(S\right))+(n\left(D\right))/(n\left(S\right))-(n\left(C\cap D\right))/(n\left(S\right)) \end{gathered}`

Here,

`\begin{gathered} n\left(C\right)=n\left(green\text{ balls}\right) \\ =5 \\ n\left(D\right)=n\left(balls\text{ numbered less than 5}\right) \\ =4 \\ n\left(C\cap D\right)=n\left(a\text{ }green\text{ }ball\text{ }or\text{ }a\text{ }ball\text{ }numbered\text{ }less\text{ }than\text{ }5\right) \\ =0 \end{gathered}`

Then,

`\begin{gathered} P\left(C\cup D\right)=(5)/(18)+(4)/(18)-0 \\ =(9)/(18) \\ =(1)/(9) \\ =0.11 \end{gathered}`

Hence, the probabilty of getting a green ball or a ball numbered less than 5 is 0.11.