# If I flip a nonbiased coin 13 times what is the probability that I get tails 10 times?

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

• How many possible outcomes are there? How many ways are there to choose the $10$ positions of the tails (or the $3$ positions of the heads) out of the $13$ flips? What have you tried on this problem? – Michael Burr Feb 1 '17 at 19:09
• Edited the question – Hybrid Feb 1 '17 at 19:11
• exactly 10 tails or at least 10 tails? – Kiran Feb 1 '17 at 19:12
• Fixed again, thanks – Hybrid Feb 1 '17 at 19:15

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

$$\dbinom{13}{10}\left(\dfrac{1}{2}\right)^{10}\left(\dfrac{1}{2}\right)^{3}+\dbinom{13}{11}\left(\dfrac{1}{2}\right)^{11}\left(\dfrac{1}{2}\right)^{2}+\dbinom{13}{12}\left(\dfrac{1}{2}\right)^{12}\left(\dfrac{1}{2}\right)^{1}+\dbinom{13}{13}\left(\dfrac{1}{2}\right)^{13}\left(\dfrac{1}{2}\right)^{0}\\=\left(\dfrac{1}{2}\right)^{13}\left[\dbinom{13}{10}+\dbinom{13}{11}+\dbinom{13}{12}+\dbinom{13}{13}\right]$$

If you want exact 10 tails then the possibility is: $$\binom{13}{10}\left(\frac{1}{2}\right)^{10}\left(\frac{1}{2}\right)^{13-10}=\frac{13*12*11}{3*2*1}\left(\frac{1}{2}\right)^{13}$$

where $\binom{13}{10}$ find all possible combination.