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Four coins are tossed simultaneously .what is the probability of getting exactly $2$ heads

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Toolbox:
  • A random variable $X$ is said to follow a binomial distribution of its probability mass function is given by
  • $P(X=x)=p(x)=\left\{\begin{array}{1 1}nC_xp^xq^{n-x},&x=0,1.......n\\0,&otherwise\end{array}\right.$
  • Constants of Binomial Distribution :
  • Mean = np
  • Variance = npq
  • Standard deviation =$\sqrt{ variance }=\sqrt{ npq}$
  • In a bionomial distribution Mean > Variance.
  • The parameters of the distribution are $n,p\quad X\sim B(n,p)$
Step 1:
Let $X$ be the random variable denoting the number of times heads turns up when 4 coins are tossed simultaneously.
The probability of getting a head in 1 throw =$\large\frac{1}{2}$$=p$
$X\sim B(4,\large\frac{1}{2})$
Step 2:
The probability distribution of $X$ is given by
$P(X=x)=nC_xp^xq^{n-x}$
$P(X=x)=4C_x(\large\frac{1}{2})^x(\large\frac{1}{2})^{4-x}$
$\qquad\qquad=(\large\frac{1}{2})^4$$4C_x\qquad x=0,1,2,3,4$
Step 3:
$P(X=2)=(\large\frac{1}{2})^4$$4C_2$
$\qquad\qquad=\large\frac{1}{16}\times \frac{4\times 3}{1\times 2}$
$\qquad\qquad=\large\frac{3}{8}$
answered Sep 18, 2013 by sreemathi.v
 

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