Question

Cards are drawn one by one at random from a well-shuffled full pack of 52 cards until 2 aces are obtained for the first time. If N is the number of cards required to be drawn, then what should be the probability of obtaining 2 aces when N is equal to n.
 \[\dfrac{{\left[ {\left( {n - 1} \right)\left( {52 - n} \right)\left( {51 - n} \right)} \right]}}{{\left[ {50 \times 49 \times 17 \times 13} \right]}}\]
 \[\dfrac{{\left[ {2\left( {n - 1} \right)\left( {52 - n} \right)\left( {51 - n} \right)} \right]}}{{\left[ {50 \times 49 \times 17 \times 13} \right]}}\]
 \[\dfrac{{\left[ {3\left( {n - 1} \right)\left( {52 - n} \right)\left( {51 - n} \right)} \right]}}{{\left[ {50 \times 49 \times 17 \times 13} \right]}}\]
 \[\dfrac{{\left[ {4\left( {n - 1} \right)\left( {52 - n} \right)\left( {51 - n} \right)} \right]}}{{\left[ {50 \times 49 \times 17 \times 13} \right]}}\]

Answer 1

Hint: In this question we will use the conditional probability concept. It is given in the question that cards are drawn till 2 aces are obtained for the first time. This indicates that ${n^{th}}$ card has to be an ace. Another ace has already been drawn from the first $\left( {n - 1} \right)$ number of cards. Thus first find the probability of drawing first $\left( {n - 1} \right)$ number of cards given that there is one ace drawn necessarily and then find the probability of ${n^{th}}$ card being an ace. Finally multiply the two probabilities to find the final answer.


Complete step by step solution: 
Since cards are drawn till 2 aces are obtained for the first time, therefore ${n^{th}}$ card is an ace.
We will first find the probability of drawing the first $\left( {n - 1} \right)$ number of cards having one and only one ace necessarily. Let us call this even A so its probability can be written as $P(A)$ .
We know that there are 4 aces in a deck of cards (having a total 52 number of cards) i.e., 48 non-ace cards.
So the probability of drawing the first ace will be $^4{C_1}$ .
Now 1 card has been drawn from the $\left( {n - 1} \right)$ cards so now we are left with the $\left( {n - 2} \right)$ number of non-ace cards. Therefore the probability of drawing $\left( {n - 2} \right)$ non- ace cards out of 48 non-ace cards will be \[\] $^{48}{C_{(n - 2)}}$ .
Also, total number of ways of drawing $\left( {n - 1} \right)$ cards from 52 cards is $^{52}{C_{(n - 1)}}$
Therefore, \[P(A) = \dfrac{{^4{C_1}{ \times ^{48}}{C_{(n - 2)}}}}{{^{52}{C_{(n - 1)}}}}\] . ...(1)
Now the probability of drawing an ace card when one ace has already been drawn is $^3{C_1}$ . Let this event be called B.
This implies that $P(B){ = ^3}{C_1}$ .
The probability of the event B happening when event A has already occurred can be written as $P\left( {B/A} \right)$ i.e., probability of drawing second ace at ${n^{th}}$ number when $(n - 1)$ cards have already been drawn having only one ace.
Here, $P(B/A) = \dfrac{{^3{C_1}}}{{\left[ {52 - (n - 1)} \right]}}$ .
Here we have written \[\left[ {52 - \left( {n - 1} \right)} \right]\] because $(n - 1)$ cards have already been drawn from 52 cards and now we are left with \[\left[ {52 - \left( {n - 1} \right)} \right]\] cards only.
Thus, $P(B/A) = \dfrac{{^3{C_1}}}{{\left( {53 - n} \right)}}$ ...(2)
Using the concept of conditional probability we know that, $P\left( {B/A} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P(A)}}$ .
This implies that \[P\left( {A \cap B} \right) = P\left( {B/A} \right) \times P(A)\] ...(3)
 where \[P\left( {A \cap B} \right)\] is the required probability to be found according to the question i.e., probability of obtaining 2 aces when N is equal to n.
Substitute (1) and (2) in equation (3), we get
 \[P\left( {A \cap B} \right) = \dfrac{{^3{C_1}}}{{\left( {53 - n} \right)}} \times \dfrac{{^4{C_1}{ \times ^{48}}{C_{(n - 2)}}}}{{^{52}{C_{(n - 1)}}}}\]
Using the formula of combination, $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$ .
This implies that \[P\left( {A \cap B} \right) = \dfrac{{\dfrac{{3!}}{{1!(3 - 1)!}}}}{{\left( {53 - n} \right)}} \times \dfrac{{\dfrac{{4!}}{{1!(4 - 1)!}} \times \dfrac{{48!}}{{(n - 2)!(48 - (n - 2))!}}}}{{\dfrac{{52!}}{{(n - 1)!(52 - (n - 1))!}}}}\]
 \[P\left( {A \cap B} \right) = \dfrac{3}{{\left( {53 - n} \right)}} \times \dfrac{{4 \times \dfrac{{48!}}{{(n - 2)!(50 - n)!}}}}{{\dfrac{{52 \times 51 \times 59 \times 49 \times 48!}}{{(n - 1)(n - 2)!(53 - n)(52 - n)(51 - n)(50 - n)!}}}}\]
Solving further, we get
 \[P\left( {A \cap B} \right) = \dfrac{{\left( {n - 1} \right)}}{{\dfrac{{13 \times 50 \times 17 \times 49}}{{(52 - n)(51 - n)}}}}\]
Hence, \[P\left( {A \cap B} \right) = \dfrac{{\left( {n - 1} \right)(52 - n)(51 - n)}}{{13 \times 17 \times 49 \times 50}}\] .
Hence, the correct option is A.

Note: The words “until 2 aces are obtained for the first time” from the question does not mean that the two aces should be drawn together i.e., one after the other. Other cards can be drawn between 2 ace cards but ${n^{th}}$ card has to be an ace. Also, one should not confuse between $P(B)$ and $P(B/A)$ . $P(B)$ is probability of B whereas $P(B/A)$ is probability of B given that A has already occurred i.e., conditional probability.