Answer:
A)[tex]P(A\cap B)=\frac{1}{3}[/tex]
B)[tex]P(A)=\frac{1}{3}[/tex]
C)[tex]P(A\mid B)=\frac{1}{2}[/tex]
D)[tex]P(B)=\frac{2}{3}[/tex]
Step-by-step explanation:
As shown in Venn diagram :
We can see that there are two outcomes that comes in the intersection part of A and B i.e. 1 and 2.
Total number of outcomes =6
So, [tex]P(A\cap B)=\frac{2}{6}=\frac{1}{3}[/tex]
Next, As we can see that there are only two outcomes in the part of A i.e.
[tex]P(A)=\frac{3}{6}=\frac{1}{3}[/tex]
Now, we have to find [tex]P(A\mid B)[/tex]
AS we know the formula for "Conditional Probability":
[tex]PA\mid B)=\frac{P(A\cap B)}{P(B)}\\\\PA\mid B)=\frac{\frac{2}{6}}{\frac{4}{6}}\\\\PA\mid B)=\frac{2}{4}=\frac{1}{2}[/tex]
Now, lat but not the least we will find P(B), as there are 4 outcomes we can see in the section of B i.e. 3, 4, 5, 6.
[tex]P(B)=\frac{4}{6}=\frac{2}{3}[/tex]
Hence,
A)[tex]P(A\cap B)=\frac{1}{3}[/tex]
B)[tex]P(A)=\frac{1}{3}[/tex]
C)[tex]P(A\mid B)=\frac{1}{2}[/tex]
D)[tex]P(B)=\frac{2}{3}[/tex]