A) \[\frac{9}{70}\]
B) \[\frac{35}{91}\]
C) \[\frac{9}{91}\]
D) \[\frac{9}{35}\]
Correct Answer: D
Solution :
[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 \[=\lambda {{(3)}^{2}}=\frac{9}{91}\] |
| \[p(B/A)=\frac{(A\cap B)}{P\,(A)}=\frac{P\,(B)}{P\,(A)}\] \[=\frac{9/91}{35/91}=9/35\] |
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