A) \[\frac{2}{3}\]
B) \[\frac{1}{3}\]
C) \[\frac{2}{9}\]
D) None of these
Correct Answer: A
Solution :
Probability of getting 13 red cards is P (R) = last card is black than red card is selected + lost card is red than red card is selected \[P(R)=P(LB)\cdot P\left( \frac{P}{LB} \right)+P(LR)P\left( \frac{R}{LR} \right)\] \[P(R)=\frac{26}{52}\cdot \frac{^{26}{{C}_{13}}}{^{51}{{C}_{3}}}+\frac{26}{52}\cdot \frac{^{25}{{C}_{13}}}{^{51}{{C}_{13}}}\] \[P(R)=\frac{1}{2}\cdot \frac{^{26}{{C}_{13}}}{^{51}{{C}_{3}}}+\frac{1}{2}\cdot \frac{^{25}{{C}_{13}}}{^{51}{{C}_{13}}}\] \[P\left( \frac{LB}{R} \right)=\frac{P(LB).P\left( \frac{R}{LB} \right)}{P(R)}\] \[=\frac{\frac{1}{2}\cdot \frac{^{26}{{C}_{13}}}{^{51}{{C}_{13}}}}{\frac{1}{2}.\frac{^{26}{{C}_{13}}}{^{51}{{C}_{13}}}+\frac{1}{3}\cdot \frac{^{25}{{C}_{13}}}{^{51}{{C}_{13}}}}=\frac{2}{3}\]
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