$\begin{array}{1 1}\text{A and B are independent events} \\\text{A and B are not independent events} \end{array} $

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- If A and B are independant events, \(P(A\cap\;B)=P(A)\;P(B)\)

The sample space for a coin toss and a roll of die can be expressed as follows:

S = $\begin{Bmatrix} 1H & 2H & 3H & 4H&5H &6H \\ 1T & 2T & 3T& 4T& 5T & 6T \end{Bmatrix}$

Let A be the event where a head appears.

A = $\begin{Bmatrix} 1H & 2H & 3H & 4H&5H &6H \\ \end{Bmatrix}$

Then P(A) = $\large \frac{6}{12} = \frac{1}{2}$

Let B be the event that a 3 appears on the die.

B = $\begin{Bmatrix} 3H & 3T \\ \end{Bmatrix}$

Then P(B) = $\large\frac{3}{12} = \frac{1}{6}$

We can see that A $\cap$ B = $\begin{Bmatrix} 3H\\ \end{Bmatrix}$

Therefore, P (A $\cap$ B) = $\large\frac{1}{12}$

We know that if A and B are independant events, \(P(A\cap\;B)=P(A)\;P(B)\)

$\Rightarrow P(A) \; P(B) = \large \frac {1}{2}$$ \times \large \frac{1}{6} = \frac{1}{12} =$ P(A $\cap$ B).

Therefore A and B are independent events.

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