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Q. 8 Suppose X has a binomial distribution $B\left [ 6,\frac{1}{2} \right ].$ Show that $X=3$ is the most likely outcome.

(Hint : $P(X=3)$ is the maximum among all of $P(x_{i})$ , $x_{i}=0,1,2,3,4,5,6$ )

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X is a random variable whose binomial distribution is $B\left [ 6,\frac{1}{2} \right ].$

Here , n=6 and $P=\frac{1}{2}$ .

$\therefore \, \,q=1-P= 1-\frac{1}{2}= \frac{1}{2}$

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$=^{6}C_x .(\frac{1}{2})^{6-x}(\frac{1}{2})^x$

$=^{6}C_x (\frac{1}{2})^6$

$P(X=x)$ is maximum if $^{6}C_x$ is maximum.

$^{6}C_0 =^{6}C_6 =\frac{6!}{0!.6!}=1$

$^{6}C_1 =^{6}C_5 =\frac{6!}{1!.5!}=6$

$^{6}C_2 =^{6}C_4 =\frac{6!}{2!.4!}=15$

$^{6}C_3 =\frac{6!}{3!.3!}=20$

$^{6}C_3$ is maximum so for x=3 , $P(X=3)$ is maximum.

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seema garhwal

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