For a Binomial variate $x$, mean is 2 and variance is 1 , Then odds in favor of $X=0$ are
For a Binomial variate $x$, mean is 2 and variance is 1 , Then odds in favor of $X=0$ are
4:1
15:1
1:15
1:4
Solution
Mean $=n p=2$, variance $=n p q=1$
$\Rightarrow q=\frac{1}{2}, p=\frac{1}{2}$ and $n=4$
Now $P(x=0)={ }^4 C_0 P^0 \cdot q^4=1 \times 1 \times\left(\frac{1}{2}\right)^4=\frac{1}{16}$
$\Rightarrow$ odds in favour of $p(x=0)=\frac{\frac{1}{16}}{1-\frac{1}{16}}=1: 15$