Probability 59493

Determine the probability of a random event out of 10 randomly selected bridge cards. There will be at least three aces. Note This is a team game, with 52 cards in the deck, of which four aces.

Correct answer:

p =  0.0194

Step-by-step explanation:

$$\begin{gathered} C_7(48)=\left(\genfrac{}{}{0ex}{}{48}{7}\right)=\frac{48!}{7!(48-7)!}=\frac{48\cdot 47\cdot 46\cdot 45\cdot 44\cdot 43\cdot 42}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=73629072 \\ {C}_6(48)=\left(\genfrac{}{}{0ex}{}{48}{6}\right)=\frac{48!}{6!(48-6)!}=\frac{48\cdot 47\cdot 46\cdot 45\cdot 44\cdot 43}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=12271512 \\ {C}_3(4)=\left(\genfrac{}{}{0ex}{}{4}{3}\right)=\frac{4!}{3!(4-3)!}=\frac{4}{1}=4 \\ {C}_{10}(52)=\left(\genfrac{}{}{0ex}{}{52}{10}\right)=\frac{52!}{10!(52-10)!}=15820024220 \\ {p}_3=\frac{\left(\genfrac{}{}{0ex}{}{4}{3}\right)\cdot \left(\genfrac{}{}{0ex}{}{52-4}{10-3}\right)}{\left(\genfrac{}{}{0ex}{}{52}{10}\right)}=\frac{4\cdot \left(\genfrac{}{}{0ex}{}{52-4}{10-3}\right)}{15820024220}=\frac{144}{7735}\doteq 0.0186 \\ {C}_4(4)=\left(\genfrac{}{}{0ex}{}{4}{4}\right)=\frac{4!}{4!(4-4)!}=\frac{1}{1}=1 \\ {p}_4=\frac{\left(\genfrac{}{}{0ex}{}{4}{4}\right)\cdot \left(\genfrac{}{}{0ex}{}{52-4}{10-4}\right)}{\left(\genfrac{}{}{0ex}{}{52}{10}\right)}=\frac{1\cdot \left(\genfrac{}{}{0ex}{}{52-4}{10-4}\right)}{15820024220}=\frac{6}{7735}\doteq 0.0008 \\ p=p_3+p_4=0.0186+0.0008=\frac{30}{1547}=0.0194 \end{gathered}$$

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