Question

Cards are drawn one-by-one at random from a well-shuffled pack of 52 playing cards until 2 aces are obtained from the first time. The probability that 18 draws are required for this is
A. \[\dfrac{3}{34}\]
B. \[\dfrac{17}{455}\]
C. \[\dfrac{561}{15925}\]
D. None of these.

Answer 1

Hint: From the given data, we can understand that the eighteenth card is the second ace. So from the first 17 draws, we must get exactly one ace. We know that a pack of 52 cards contain 4 aces. One ace appears in the first 17 draws. The remaining non ace cards are $52-4=48$ and the number of draws in which non-aces appear is $17-1=16$ . Hence, we can write the probability of getting 1 ace and 16 non-aces as $P=\dfrac{^{4}{{C}_{1}}{{\times }^{48}}{{C}_{16}}}{^{52}{{C}_{17}}}$ . Now, we have multiplied this by the probability of obtaining a second ace in the next draw , that is, eighteenth draw. This probability is given as $P=\dfrac{3}{35}$ . Hence, we can write the required probability as $P=\dfrac{^{4}{{C}_{1}}{{\times }^{48}}{{C}_{16}}}{^{52}{{C}_{17}}}\times \dfrac{3}{35}$ .

Complete step by step answer:
It is given that cards are drawn one-by-one at random from a well-shuffled pack of 52 playing cards until 2 aces are obtained from the first time. We have to find the probability that 18 draws are required for this.
From the given data, we can understand that the eighteenth card is the second ace. So from the first 17 draws, we must get exactly one ace.
We know that a pack of 52 cards contain 4 aces. One ace appears in first 17 draws. The remaining non ace cards are $52-4=48$ and the number of draws in which non-aces appear is $17-1=16$ . Hence, we can write the probability of getting 1 ace and 16 non-aces as
$P=\dfrac{^{4}{{C}_{1}}{{\times }^{48}}{{C}_{16}}}{^{52}{{C}_{17}}}$
Now, we have multiply this by the probability of obtaining a second ace in the next draw, that is, eighteenth draw. This probability is given as
$P=\dfrac{3}{35}$
3 is the remaining aces , that is, $4-1$ , and 35 is the remaining cards, that is, $52-17=35$ .
Hence, we can write the required probability as
$P=\dfrac{^{4}{{C}_{1}}{{\times }^{48}}{{C}_{16}}}{^{52}{{C}_{17}}}\times \dfrac{3}{35}$
We know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and \[^{n}{{C}_{1}}=n\] .
Hence, we can write the above probability as
$P=\dfrac{4\times \dfrac{48!}{16!\left( 48-16 \right)!}}{\dfrac{52!}{17!\left( 52-17 \right)!}}\times \dfrac{3}{35}$
Let us solve this.
$\begin{align}
  & P=\dfrac{4\times \dfrac{48!}{16!32!}}{\dfrac{52!}{17!35!}}\times \dfrac{3}{35} \\
 & \Rightarrow P=4\times \dfrac{48!}{16!32!}\times \dfrac{17!35!}{52!}\times \dfrac{3}{35} \\
\end{align}$
Let us expand the factorial.

$P=4\times \dfrac{48!}{16!32!}\times \dfrac{17\times 16!\times 35\times 34\times 33\times 32!}{52\times 51\times 50\times 49\times 48!}\times \dfrac{3}{35}$
Let us cancel common terms. We will get
$\begin{align}
  & P=4\times \dfrac{17\times 34\times 33}{52\times 51\times 50\times 49}\times 3 \\
 & \Rightarrow P=\dfrac{17\times 34\times 33}{13\times 17\times 50\times 49} \\
\end{align}$
Now, let us cancel the common factors from numerator and denominator. We will get
$P=\dfrac{17\times 33}{13\times 25\times 49}$
$P=\dfrac{561}{15925}$

So, the correct answer is “Option C”.

Note: You must know the number of each type of card in a pack of 52 cards. To get the required probability, you may make mistakes by adding the probability of getting the second ace on eighteenth draw and the probability of getting 1 ace and 16 non-aces. You may write \[^{n}{{C}_{1}}\] as 1.