A) \[\frac{7}{13}\]
B) \[\frac{3}{26}\]
C) \[\frac{63}{221}\]
D) \[\frac{55}{221}\]
E) \[\frac{5}{7}\]
Correct Answer: D
Solution :
\[n(S)={{\,}^{52}}{{C}_{2}}=\frac{52\times 51}{2}=1326\] Let\[{{E}_{1}}=\] event of getting both red cards and \[{{E}_{2}}=\] = event of getting both Kings Then,\[{{E}_{1}}\cap {{E}_{2}}=\]event of getting two King of red cards \[n({{E}_{1}})={{\,}^{26}}{{C}_{2}}=\frac{26\times 25}{2}=325,\] \[n({{E}_{2}})=\frac{4\times 3}{2\times 1}=6\] \[n({{E}_{1}}\cap {{E}_{2}})={{\,}^{2}}{{C}_{2}}=1\] Now, \[P({{E}_{1}})=\frac{n({{E}_{1}})}{n(S)}=\frac{325}{1326},P({{E}_{2}})=\frac{n({{E}_{2}})}{n(S)}=\frac{6}{1326}\] \[\therefore \]\[P({{E}_{1}}\cap {{E}_{2}})=\frac{1}{1326}\] P(Bothe red both Kings) \[=P({{E}_{1}}\cup {{E}_{2}})=P({{E}_{1}})+P({{E}_{2}})-P({{E}_{1}}\cap {{E}_{2}})\] \[=\left( \frac{325}{1376}+\frac{6}{1328}-\frac{1}{1326} \right)=\frac{55}{221}\]
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