Flipping a coin is an independent event, and has a chance of 50% of either heads or tails. So lets say that I flip a coin 13 times, what is the probability that I get 10 tails in any order/any number of possible outcomes, in 13 flips?
Edit: The probability of at least ten tails
Best Answer
Let's first find probability for exactly $10$ tails
Select any $10$ tosses: $\dbinom{13}{10}$
Each of these $10$ tosses must get tail $\left(\dfrac{1}{2}\right)^{10}$
Each of these remaining $3$ tosses must get head $\left(\dfrac{1}{2}\right)^{3}$
So, probability is
$\dbinom{13}{10}\left(\dfrac{1}{2}\right)^{10}\left(\dfrac{1}{2}\right)^{3}$
Similarly we can find probability for $11, 12, 13$ heads also.
So, required probability is
$$\dbinom{13}{10}\left(\dfrac{1}{2}\right)^{10}\left(\dfrac{1}{2}\right)^{3}+\dbinom{13}{11}\left(\dfrac{1}{2}\right)^{11}\left(\dfrac{1}{2}\right)^{2}+\dbinom{13}{12}\left(\dfrac{1}{2}\right)^{12}\left(\dfrac{1}{2}\right)^{1}+\dbinom{13}{13}\left(\dfrac{1}{2}\right)^{13}\left(\dfrac{1}{2}\right)^{0}\\=\left(\dfrac{1}{2}\right)^{13}\left[\dbinom{13}{10}+\dbinom{13}{11}+\dbinom{13}{12}+\dbinom{13}{13}\right]$$