It is given that when head appears on the coin, the event is A and when 3 on the die, the event is B.
Total possible outcomes are 12. Sample space is given by,
S={ ( H,1 ),( H,2 ),( H,3 ),( H,4 ),( H,5 ),( H,6 ), ( T,1 ),( T,2 ),( T,3 ),( T,4 ),( T,5 ),( T,6 ) }
The event A can be possible as,
A={ ( H,1 ),( H,2 ),( H,3 ),( H,4 ),( H,5 ),( H,6 ) }
Then, the probability of A is,
P( A )= 6 12 = 1 2
The event B can be possible as,
B={ ( H,3 ),( T,3 ) }
Then, the probability of B is,
P( B )= 2 12 = 1 6
The condition when the head appear on the coin and 3 on the die is ( H,3 ).
The probability is given as,
P( A∩B )= 1 12 (1)
The formula when two events are independents is given as,
P( A∩B )=P( A )⋅P( B )
Substitute the value of P( A )= 1 2 and P( B )= 1 6 in the above formula
P( A )×P( B )= 1 2 × 1 6 P( A )×P( B )= 1 12 (2)
Equation (1) and (2) P( A∩B )=P( A )⋅P( B ).
Hence, A and B are independent events.