A random variable X has the following probability distribution: X 1 2…

The correct answer is Option (4) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

$\text{Given distribution:}$

$P(1)=k,\;P(2)=2k,\;P(3)=2k,\;P(4)=3k,\;P(5)=k^2,\;P(6)=2k^2,\;P(7)=7k^2+k.$

$\sum P(X)=1.$

$k+2k+2k+3k+k^2+2k^2+(7k^2+k)=1.$

$9k+10k^2=1.$

$10k^2+9k-1=0.$

$k=\frac{-9+\sqrt{81+40}}{20}=\frac{-9+11}{20}=\frac{1}{10}.$

$(A)\;k=\frac{1}{10}\Rightarrow(III).$

$(B)\;P(X<3)=P(1)+P(2)=k+2k=3k=\frac{3}{10}\Rightarrow(IV).$

$(C)\;P(X>2)=1-[P(1)+P(2)]=1-3k=1-\frac{3}{10}=\frac{7}{10}\Rightarrow(I).$

$(D)\;P(2<\text{ X }<7)=P(3)+P(4)+P(5)+P(6).$

$=2k+3k+k^2+2k^2=5k+3k^2.$

$=5\cdot\frac{1}{10}+3\cdot\frac{1}{100}=\frac{1}{2}+\frac{3}{100}=\frac{53}{100}\Rightarrow(II).$

$(A)\rightarrow(III),\;(B)\rightarrow(IV),\;(C)\rightarrow(I),\;(D)\rightarrow(II).$