Question

All the face cards from a pack of 52 playing cards are removed. From the remaining pack half of the cards are randomly removed without looking at them and then randomly drawn two cards simultaneously from the remaining. If the probability that, two cards drawn are both aces, is $\frac{p\left({}^{38}{C}_{20}\right)}{{}^{40}{C}_{20}{}^{20}{C}_2}$, find p.

Answer 1

Answer: 6

$52\stackrel{\text{ face card removed }}{\to }40\stackrel{20\text{ drawn randomly }}{\to }$
Let $E_0:20\text{ cards randomly removed has no aces. }$
$E_1:20\text{ cards randomly removed has exactly one ace. }$
$E_2:20\text{ cards randomly removed has exactly }2\text{ aces. }$
$E:\text{ event that }2\text{ drawn from the remaining }20\text{ cards has both the aces. }$
$P(E)=P\left(E\cap E_0\right)+P\left(E\cap E_1\right)+P\left(E\cap E_2\right)$
$=P\left(E_0\right)\cdot P\left(E/E_0\right)+P\left(E_1\right)\cdot P\left(E/E_1\right)+P\left(E_2\right)\cdot P\left(E/E_2\right)$


$=\frac{{}^4{C}_0\cdot {}^{36}{C}_{20}}{{}^{40}{C}_{20}}\cdot \frac{{}^4{C}_2}{{}^{20}{C}_2}+\frac{{}^4{C}_1\cdot {}^{36}{C}_{19}}{{}^{40}{C}_{20}}\cdot \frac{{}^3{C}_2}{{}^{20}{C}_2}+\frac{{}^4{C}_2\cdot {}^{36}{C}_{18}}{{}^{40}{C}_{20}}\cdot \frac{{}^2{C}_2}{{}^{20}{C}_2}$
$=\frac{{}^{36}{C}_{20}\cdot {}^4{C}_2+{}^4{C}_1\cdot {}^{36}{C}_{19}\cdot {}^3{C}_2+{}^4{C}_2\cdot {}^{36}{C}_{18}\cdot {}^2{C}_2}{{}^{40}{C}_{20}\cdot {}^{20}{C}_2}$
$=\frac{6^{36}{C}_{20}+12\cdot {}^{36}{C}_{19}+6\cdot {}^{36}{C}_{18}}{{}^{40}{C}_{20}\cdot {}^{20}{C}_2}=\frac{6\left[{}^{36}{C}_{20}+{}^{36}{C}_{19}+{}^{36}{C}_{19}+{}^{36}{C}_{18}\right]}{{}^{40}{C}_{20}\cdot {}^{20}{C}_2}$
$=\frac{6\left({}^{37}{C}_{20}+{}^{37}{C}_{19}\right)}{{}^{40}{C}_{20}\cdot {}^{20}{C}_2}=\frac{6\left({}^{38}{C}_{20}\right)}{{}^{40}{C}_{20}\cdot {}^{.20}{C}_2}\Rightarrow p=6$