# If X is the number of tails in three tosses of a coin ,determine the standard deviation of X.

Toolbox:
• In a toss of 3 coin .X denotes the number of tails
• X can take value X=0(notail);(X=1)(1 tail)
• X=2(2tails);X=3(3tails)
• $$x_i$$=i-1,2,3,
• $$P_i$$=are there correrpending proabilities
• =$$\sigma\;\Large\;\sqrt{\sum_iP_i\;x_i^2\;-\;\sum_i\;P_i\;x_i)^2}$$
• Probability of getting 1 head in a single throw=$$\large\frac{1}{2}$$;tail=$$\large\frac{1}{2}$$
P(x=0)=P(no tail)=$$P(H\;H\;H)$$
=$$\Large\frac{1}{2}\;\times\;\frac{1}{2}\;\times\;\frac{1}{2}\;=\;\frac{1}{8}$$
P(X=1)=P(1tail)
$$P(T\;T\:H\:,T\;H\;H\;,H\;T\;T)$$

=$$\Large\frac{1}{8}\;+\;\frac{1}{8}\;+\;\frac{1}{8}\;=\;\frac{3}{8}$$
P(X=2)=P(2tail)
$$P(H\;T\;T\;,T\;H\;T\;,T\;T\;H)$$
$$\Large\frac{1}{8}\;+\;\frac{1}{8}\;+\;\frac{1}{8}\;=\;\frac{3}{8}$$

P(X=3)=P(3tail)
$$P(T\;T\;T)$$
=$$\Large\;\frac{1}{8}$$
$$\Large\;\sum\;P_iX-i\;=\;\frac{1}{8}\;\times0+\;\frac{3}{8}\;\times1+\;\frac{3}{8}\;\times2+\;\frac{1}{8}\;\times3$$
$$\Large\;\frac{12}{8}$$
$$\Large\;\sum\;P_i\;X_i^2\;=\;\frac{1}{8}\;\times0+\;\frac{3}{8}\;\times1+\;\frac{3}{8}\;\times4+\;\frac{1}{8}\;\times9$$
$$\Large\frac{42}{8}$$
$$\Large\;\sigma=\sqrt{\frac{42}{8}-(\frac{12}{8})^2}$$=$$\Large\sqrt\frac{192}{64}$$
$$\Large\frac{\sqrt3}{2}$$

edited Jun 4, 2013