Answer:
Once the number of possible heads and tail in 12 tosses is [tex]2^{12} =4096[/tex]
[tex]\frac{12!}{\left(7!\right)\left(5!\right)}\\\\\frac{12\cdot \:11\cdot \:10\cdot \:9\cdot \:8}{5!}\\\\\frac{95040}{120}\\\\792[/tex]
So the probability of getting exactly 8 heads in 12 coin tosses is: [tex]\frac{792}{4096}[/tex]
