JAMIA 2022 — Mathematics PYQ

1. Define the Events

• Event $E_1$ (Die A shows 4):

  The outcomes are $(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$.

  $$P(E_1)=\frac{6}{36}=\frac{1}{6}$$

• Event $E_2$ (Die B shows 2):

  The outcomes are $(1,2),(2,2),(3,2),(4,2),(5,2),(6,2)$.

  $$P(E_2)=\frac{6}{36}=\frac{1}{6}$$

• Event $E_3$ (Sum is odd):

  A sum is odd if one die is even and the other is odd. There are 18 such outcomes.

  $$P(E_3)=\frac{18}{36}=\frac{1}{2}$$

2. Check Pairwise Independence

Two events $X$ and $Y$ are independent if $P(X\cap Y)=P(X)\cdot P(Y)$.

• For $E_1$ and $E_2$:

  $E_1\cap E_2$ is the outcome $(4,2)$.

  $$P(E_1\cap E_2)=\frac{1}{36}$$
  $$P(E_1)\cdot P(E_2)=\frac{1}{6}\cdot \frac{1}{6}=\frac{1}{36}$$

  Since $1/36=1/36$, $E_1$ and $E_2$ are independent.

• For $E_2$ and $E_3$:

  $E_2\cap E_3$ means Die B is 2 (even) and the sum is odd. This implies Die A must be odd $(1,2),(3,2),(5,2)$.

  $$P(E_2\cap E_3)=\frac{3}{36}=\frac{1}{12}$$
  $$P(E_2)\cdot P(E_3)=\frac{1}{6}\cdot \frac{1}{2}=\frac{1}{12}$$

  Since $1/12=1/12$, $E_2$ and $E_3$ are independent.

• For $E_1$ and $E_3$:

  $E_1\cap E_3$ means Die A is 4 (even) and the sum is odd. This implies Die B must be odd $(4,1),(4,3),(4,5)$.

  $$P(E_1\cap E_3)=\frac{3}{36}=\frac{1}{12}$$
  $$P(E_1)\cdot P(E_3)=\frac{1}{6}\cdot \frac{1}{2}=\frac{1}{12}$$

  Since $1/12=1/12$, $E_1$ and $E_3$ are independent.

3. Check Mutual Independence

Events are mutually independent if $P(E_1\cap E_2\cap E_3)=P(E_1)\cdot P(E_2)\cdot P(E_3)$.

• $E_1\cap E_2\cap E_3$: Die A is 4, Die B is 2, and the sum ($4+2=6$) is odd.

• Since the sum 6 is even, this event is impossible.

  $$P(E_1\cap E_2\cap E_3)=0$$

Now calculate the product:

Since $0\ne \frac{1}{72}$, the events are not mutually independent.

Final Answer

The statements regarding pairwise independence ($E_1,E_2$, $E_2,E_3$, and $E_1,E_3$) are all True.

The statement "$E_1,E_2,E_3$ are mutually independent" is False.