Question

The Venn diagram shows the results of two events resulting from rolling a number cube.

Answer 1

Answer:  

A)$P(A\cap B)=\frac{1}{3}$

B)$P(A)=\frac{1}{3}$

C)$P(A\mid B)=\frac{1}{2}$

D)$P(B)=\frac{2}{3}$

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, $P(A\cap B)=\frac{2}{6}=\frac{1}{3}$

Next, As we can see that there are only two outcomes in the part of A i.e.

$P(A)=\frac{3}{6}=\frac{1}{3}$

Now, we have to find $P(A\mid B)$

AS we know the formula for "Conditional Probability":

$PA\mid B)=\frac{P(A\cap B)}{P(B)}\\\\PA\mid B)=\frac{\frac{2}{6}}{\frac{4}{6}}\\\\PA\mid B)=\frac{2}{4}=\frac{1}{2}$

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.

$P(B)=\frac{4}{6}=\frac{2}{3}$

Hence,

A)$P(A\cap B)=\frac{1}{3}$

B)$P(A)=\frac{1}{3}$

C)$P(A\mid B)=\frac{1}{2}$

D)$P(B)=\frac{2}{3}$

Answer 2

Answer:   $P(A\cap B)=\frac{1}{3}$

$P(A)=\frac{1}{3}$

$P(B)=1$

$P(A|B)=\frac{1}{3}$

Step-by-step explanation:

From the given figure it can be seen that

Total number on cube= n(S)=6

Intersection of A and B = n(A ∩ B)= 2

Therefore,

$P(A\cap B)=\frac{n(A\cap B)}{n(s)}=\frac{2}{6}=\frac{1}{3}$

Also, The number of elements in A = 2

Therefore,

$P(A)=\frac{n(A)}{n(s)}=\frac{2}{6}=\frac{1}{3}$

Similarly, The number of elements in B= 6

Therefore,

$P(B)=\frac{n(B)}{n(s)}=\frac{6}{6}=1$

The formula to find the conditional probability is given by :-

$P(A|B)=\frac{P(A\cap B)}{P(B)}\\\\\Rightarrow P(A|B)=\frac{\frac{1}{3}}{1}=\frac{1}{3}$