• # question_answer A die is so loaded that the probability of throwing a number is proportional to ${{K}^{2}}.$ Then the probability that the number 3 appears given that when the die is rolled the number turned up is not even is equal to A) $\frac{9}{70}$                          B) $\frac{35}{91}$ C) $\frac{9}{91}$                          D) $\frac{9}{35}$

[d]  Given, Probability of throwing a number is proportional to ${{K}^{2}}.$ $\therefore$$\sum\limits_{K=1}^{6}{\lambda {{K}^{2}}=1}$ $\Rightarrow$$\frac{\lambda (6)(6+1)(12+1)}{6}=1$$\Rightarrow$$\lambda =\frac{1}{91}$ P [a] = Probability of not getting an even number $=\lambda ({{1}^{2}}+{{3}^{2}}+{{5}^{2}})$ $=\lambda \,(35)=\frac{35}{91}$ P [b] = Probability of getting 3 $=\lambda {{(3)}^{2}}=\frac{9}{91}$ $p(B/A)=\frac{(A\cap B)}{P\,(A)}=\frac{P\,(B)}{P\,(A)}$ $=\frac{9/91}{35/91}=9/35$