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# Four coins are tossed simultaneously.what is the probability of getting at least two heads

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:
$P(X\geq 2)=P(X=2)+P(X=3)+P(X=4)$
$\qquad\qquad=(\large\frac{1}{2})^4$$[4C_2+4C_3+4C_4] \qquad\qquad=(\large\frac{1}{2})^4[\large\frac{4\times 3}{1\times 2}$$+4+1]$
$\qquad\qquad=\large\frac{1}{16}$$[11]$
$\qquad\qquad=\large\frac{11}{16}$