Comment
Share
Q)

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

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