bag full of points

We want to find P(not getting all 5 tails). So lets start by using the complement:

P(not getting all 5 tails) = 1 - P(getting all 5 tails)

Using the law of total probability we can rewrite P(getting all 5 tails) as follows:

P(getting all 5 tails) = P(all 5 tails | you picked a fair coin) * P(picking a fair coin) + P(all 5 tails | you picked the unfair coin) * P(picking the unfair coin) = (1/2)^5 * 9/10 + 1 * 1/10 = 41/320

Therefore, our final answer is 1 - 41/320 = .871875

$to\quad find\quad p(not\quad 5/5tails)\\ no\quad of\quad normal\quad coins=9\\ no.\quad of\quad unfair\quad coin\quad =1\\ total\quad no.\quad of\quad coin=10\\ \\ p(5/5\quad head)=\\ p(5head\cap normal\quad coins)+p(5head\cap unfair\quad coin)\\ p(5head\cap normal\quad coins)=p(5head|normal\quad coins)*p(normal)\\ p(5head|normal\quad coins)={ (\frac { 1 }{ 2 } }^{ 5 })*\frac { 9 }{ 10 } \\ =1/32*9/10=9/320\\ p(5head\cap unfair\quad coin)=p(5head|unfair\quad coin)*P(unfair\quad coin)\\ 1*1/10=1/10\\ p(5/5\quad head)=9/320+1/10\\ p(not\quad 5/5\quad head)=1-(9/320+1/10)\\ p(not\quad 5/5\quad head)=0.872$