Answer:
![(7)/(12)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/rlsqqbkbsw6amdr0p450qjkcesslsf3ua0.png)
Explanation:
We are given that
The probability of pulling a red marble=
![(1)/(4)](https://img.qammunity.org/2020/formulas/mathematics/high-school/iiq2xsk4vi9pqjukqb60xxgyxukyno498i.png)
The probability of pulling a blue marble=
![(1)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/o08xg954t1gbzo9avralvfomcybk63rm02.png)
We have to find the probability that a marble at randomly pulled from the urn will be one of those two colors ( red or blue)
When a marble is of red color then it can not be of blue color .
When a marble is of blue color then it can not be of red color.
![P(Red\cap blue)=0](https://img.qammunity.org/2020/formulas/mathematics/high-school/77kzgfimen0csr5rjawwitn76vzjnhhsle.png)
We know that
![P(A\cup B)=P(A)+P(B)-P(A\cap B)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hjvmv4ll3g25188919dtl3ycbphwrvckcc.png)
A=Red marble
B=Blue marble
![P(A)=(1)/(4)](https://img.qammunity.org/2020/formulas/mathematics/high-school/t6ct0yx9hyl91488bfw93dkbksibbyndn3.png)
![P(B)=(1)/(3)](https://img.qammunity.org/2020/formulas/mathematics/college/9y8ewufzpfuuno0x3gl1at6x8n9rwxgbgi.png)
Substitute the values then, we get
![P(Red\cup blue)=(1)/(4)+(1)/(3)=(3+4)/(12)=(7)/(12)](https://img.qammunity.org/2020/formulas/mathematics/high-school/zcma0fs2n7zuilf1ninf0mb1ax7qwl7prb.png)
Hence, the probability of pulling red or blue marble from the urn=
![(7)/(12)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/rlsqqbkbsw6amdr0p450qjkcesslsf3ua0.png)