# A fair coin is flipped 9 times. What is the probability of getting exactly 6 heads?

**Solution:**

The binomial distribution formula is also written in the form of n-Bernoulli trials.

**where ^{n}Cx = n! / x!(n-x)!. **

Hence,

P(x:n,p) = n! / [x! (n - x)! ]. p^{x}.(q)^{n - x}

**The tossing of a coin is a Bernoulli’s process which comprises ‘n’ Bernoulli trials and the probability of success in each trial is p and failure is q. **

q is obtained by subtracting p (probability of success) from 1.

**The probability of the desired outcome of an Bernoulli experiment which is given as follows:**

P(X = x) = \( C_{x}^{n}\textrm{}q^{_{n-x}}p^{x}, x = 0,1, 2, .....n\)

**The problem can be stated mathematically as n = 9 and x = 6 heads is**

p = 1/2 ; q = 1 - 1/2 = 1/2

**P(X = 6 heads) = \( C_{6}^{9}\textrm{}(1/2)^{_{9-6}}(1/2)^{6}, x = 0,1, 2, .....n \)**

= (84) (1/2)^{3}(1/2)^{6}

= (84) (1/8) (1/64)

= 21/128

**The probability of getting exactly 6 heads is 21/28.**

## A fair coin is flipped 9 times. What is the probability of getting exactly 6 heads?

**Summary:**

When a coin is tossed 9 times the probability of getting exactly six heads is obtained with the help of Bernoulli trials. The probability of getting exactly six heads is 21/128.