A) \[\frac{52}{169}\]
B) \[\frac{25}{169}\]
C) \[\frac{49}{169}\]
D) \[\frac{24}{169}\]
Correct Answer: B
Solution :
\[p=\frac{4}{52}=\frac{1}{13}\] \[\Rightarrow \]\[q=\frac{12}{13}\] \[n=2.p(x=k)={}^{n}{{C}_{k}}.{{q}^{x-k}}.{{p}^{k}}\] \[p(x=1)+p(x=2)={}^{2}{{C}_{1}}\frac{12}{13}.\frac{1}{13}+{}^{2}{{C}_{2}}{{\left( \frac{1}{13} \right)}^{2}}=\frac{25}{169}.\]You need to login to perform this action.
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