Coin Toss

Consider 10 coins are tossed simultaneously. What is the probability of getting at least 9 tails? Assume either heads or tails for only possible outcomes.

Hint
There are exactly two possible outcomes (heads/tails) for every coin toss. Regardless if thrown simultaneously or not, there will be a one in two chance of the coin landing tails. Thus, the total number of possible outcomes is:
$$$\frac{1}{2}\cdot \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{2^{10}}=\frac{1}{1024}$$$
Hint 2
Unlike a permutation where a particular sequence order is considered, a combination is needed to find the number of favorable events (i.e. the number of outcomes in which there are 9 or 10 tails) because all the coins are tossed simultaneously.
There are exactly two possible outcomes (heads/tails) for every coin toss. Regardless if thrown simultaneously or not, there will be a one in two chance of the coin landing tails. Thus, the total number of possible outcomes is:
$$$\frac{1}{2}\cdot \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{2^{10}}=\frac{1}{1024}$$$
Unlike a permutation where a particular sequence order is considered, a combination is needed to find the number of favorable events (i.e. the number of outcomes in which there are 9 or 10 tails) because all the coins are tossed simultaneously:
$$$C_{(n,r)}=\frac{P_{(n,r)}}{r!}=\frac{n!}{[r!(n-r)!]}$$$
where $$C_{n,r}$$ is the number of different combinations of $$n$$ distinct objects taken $$r$$ at a time, and $$P$$ is the number of different permutations. Thus, the number of favorable outcomes for our problem statement is:
$$$C_{(10,9)}+C_{(10,10)}$$$
which is the number of ways to select 9 coins out of 10 plus the number of ways to select 10 coins out of 10.
$$$=\frac{10!}{[9!(10-9)!]}+\frac{10!}{[10!(10-10)!]}=\frac{10!}{9!\cdot 1!}+\frac{10!}{10!\cdot 0!}=\frac{10}{1}+\frac{1}{1}=11$$$
Finally, the probability of getting at least 9 tails:
$$$\frac{11}{1024}$$$
$$$\frac{11}{1024}$$$