<PRE><B>Question 157543</B>
You can construct a table of all of the possible tosses.
There are 256 (2^8) possibilities.
Here&#39;e the general way to do it with 4 tosses, you can expand to 8 to get your answer.
4 tosses there are 16 possible outcomes.
 
1.H 	H 	H 	H	=4 H, 0 T 
2.H 	H 	H 	 T 	=3 H, 1 T 
3.H 	H 	 T 	H 	=3 H, 1 T 
4.H 	H 	 T 	 T 	=2 H, 2 T 
5.H 	 T 	H 	H 	=3 H, 1 T 
6.H 	 T 	H 	 T 	=2 H, 2 T 
7.H 	 T 	 T 	H 	=2 H, 2 T 
8.H 	 T 	 T 	 T 	=1 H, 3 T 
9. T 	H 	H 	H 	=3 H, 1 T 
10. T 	H 	H 	 T 	=2 H, 2 T 
11. T 	H 	 T 	H 	=2 H, 2 T 
12. T 	H 	 T 	 T 	=1 H, 3 T 
13. T 	 T 	H 	H 	=2 H, 2 T 
14. T 	 T 	H 	 T 	=1 H, 3 T 
15. T 	 T 	 T 	H 	=1 H, 3 T 
16. T 	 T 	 T 	 T 	=0 H, 4 T 
Then you can look at individual probabilities
P{4H)=1/16
P(3H)=4/16
P(2H)=6/16
P(1H)=4/16
P(0H)=1/16
P(at least 2 heads)=P(2H)+P(3H)+P(4H)=6/16+4/16+1/16=11/16
Hope it helps.
The procedure is the same for any number of tosses, it just a bit more tedious with more tosses.
Good luck. </PRE>