Question

Cards are drawn one by one at random from a well-shuffled pack of $52$ playing cards until $2$ aces are obtained for the first time. The probability that $18$ draws are required for this, is

• A. $\frac{{}^3{C}_1\times {}^{50}{C}_{16}}{{}^{52}{C}_{17}}\times \frac{{}^4{C}_1}{{}^{35}{C}_1}$
• B. $\frac{{}^3{C}_1\times {}^{47}{C}_{16}}{{}^{52}{C}_{17}}\times \frac{{}^4{C}_1}{{}^{35}{C}_1}$
• C. $\frac{{}^4{C}_1\times {}^{48}{C}_{16}}{{}^{53}{C}_{17}}\times \frac{{}^2{C}_1}{{}^{34}{C}_1}$
• D. $\frac{{}^4{C}_1\times {}^{48}{C}_{16}}{{}^{52}{C}_{17}}\times \frac{{}^3{C}_1}{{}^{35}{C}_1}$

Answer 1

The correct option is D $\frac{{}^4{C}_1\times {}^{48}{C}_{16}}{{}^{52}{C}_{17}}\times \frac{{}^3{C}_1}{{}^{35}{C}_1}$

$18$ draws are required for $2$ aces means in the first $17$ draws, there is one ace and $16$ other cards and ${18}^{\text{th}}$ draw produces an ace.
The required probability for first $17$ draws $=\frac{{}^4{C}_1\times {}^{48}{C}_{16}}{{}^{52}{C}_{17}}$
After $17$ draws, $3$ aces are remaining and $35$ cards are remaining.
The required probability for ${18}^{\text{th}}$ draw $=\frac{{}^3{C}_1}{{}^{35}{C}_1}$
$\therefore$ Required probability $=\frac{{}^4{C}_1\times {}^{48}{C}_{16}}{{}^{52}{C}_{17}}\times \frac{{}^3{C}_1}{{}^{35}{C}_1}$