A) \[P({{E}_{2}})\ge P({{E}_{1}})\]
B) \[P({{E}_{2}}\cap {{E}_{1}})=\frac{3}{10}\]
C) \[P({{E}_{2}}/{{E}_{1}})=\frac{1}{36}\]
D) \[P({{E}_{1}})=\frac{10}{3}P({{E}_{2}})\]
Correct Answer: D
Solution :
[d] \[P({{E}_{1}})=\frac{^{6}{{C}_{3}}}{216}=\frac{5}{54};P({{E}_{2}})=\frac{4+2}{216}=\frac{1}{36}\] \[{{E}_{1}}\cap {{E}_{2}}={{E}_{2}}\therefore P({{E}_{1}}\cap {{E}_{2}})=P({{E}_{2}})=1/36\] \[P({{E}_{2}}/{{E}_{1}})=\frac{P({{E}_{1}}\cap {{E}_{2}})}{P({{E}_{1}})}=\frac{1/36}{5/54}=\frac{54}{180}=\frac{3}{10}.\]
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