Question 157543: hey I will really appreciate it if you could help me with this homework question:
A coin is tossed 8 times. Calculate the probability of tossing
a) 6 heads and 2 tails
b) 7 heads and 1 tail
c) at least 6 heads
Answer by Fombitz(32388)   (Show Source):
You can put this solution on YOUR website! You can construct a table of all of the possible tosses.
There are 256 (2^8) possibilities.
Here'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.
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