GATE CSE 1996 | Question: 2.7

Question

The probability that top and bottom cards of a randomly shuffled deck are both aces is

• A. $\frac{4}{52} \times \frac{4}{52}$

• B. $\frac{4}{52} \times \frac{3}{52}$

• C. $\frac{4}{52} \times \frac{3}{51}$

• D. $\frac{4}{52} \times \frac{4}{51}$

Comments

4C2 / 52C2

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Ans is (2*4C2 * 50!) / (52!) = 12 / (52*51)

numerator is 2*4C2 * 50! as we can choose 2 aces from 4 aces in 4C2 ways and place in top and bottom position in 2 ways and then arrange the remaining 50 cards in 50! ways. 52! is for total possible arrangements.

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total 4 ace, if we get ace on either top or bottom first ,then we have to choose between 4-1 =3 aces for the remaining face card (either top or bottom depending on which is drawn first) as 1 ace is already gone.

Answer 1

There are $52$ cards including $4$ aces so the probability must be $\dfrac{4}{52}\times \dfrac{3}{51}.$

Correct Answer: $C$

Comments

If anyone doesn't know about cards, there are 4 suits (Club, Diamond, Heart, Spade) each having 13 cards (Ace,2,3,4,5,6,7,8,9, Jack, Queen, King), 26 red cards (13 diamonds + 13 hearts) and 26 black cards (13 clubs + 13 spades).

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$Ace ,2,3,4,5,6,7,8,9,10, Jack, Queen, King$

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@Peeyush Pandey Please edit your ans and include card 10 also in the 13 cards that are present in each suits as it is creating confusion. 

@pranavbhosle_ Thanks for the comment.

Answer 2

$E_1$ : first card being ace

$E_2$ : last card being ace

Note : $E_1$ and $E_2$ are dependent events (I.e) probability of last card being ace if first card is ace will be lesser than the probability of last card being ace if first card is not ace ....

So, probability of first card being ace = $4/52$

Probability of last card being ace given that first card is ace (I.e) $P(E_2/E_1)$ = $3/51$

$P$($E_1$ AND $E_2$) = $P(E_1).P(E_2/E_1)$ = $4/52 $*$ 3/51$

Correct Answer: $C$

Comments

Nice one.

We can also say that P(E1 AND E2) = P(E2).P(E1∣E2) = 4/52 * 3/51

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Bydefault events are dependent ??? Plzz clear me this

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Nice Explanation !!

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BEST explanation

Answer 3

Required Probability = Size of favourable cases / Size of total possible cases

So,

Size of set of total possible cases = No of ways we can choose aces from the deck of the cards=  52C2

 

Size of set of  favourable cases = No of ways to choose 2 Ace cards from 4 Ace cards to be placed in top and bottom of the card =  4C2

 

So the required Probability = 4C2/52C2

                                           = 4*3 / 52*51

Answer 4

The probability that the bottom card of a randomly shuffled deck is ace = 4/52

From the remaining 3 aces out of cards the probability that the top card is also an ace = 3/51.

So the required Probability =  4/52 x  3/51

 

I hope my answer helps you a lot

 

Answer 5

Total Outcomes = 52!

Favourable outcomes = 4P2 * 50!

=> Favourable outcomes/Total Outcomes = (4P2 * 50!)/52! = (4*3)/(52*51)

 

Answer C is correct.

Answer 6

Let us consider position 1 as the top and position 52 as the bottom.

For the top position (1st card), we have 4C1 ways (choosing any one of the 4 Aces).

For the bottom position (52nd card), we have 3C1 ways (since one Ace is already used for the top, only 3 Aces remain).

For the middle 50 positions, we can arrange the remaining cards in 50! ways.

Therefore, total favourable cases = (4C1) × (3C1) × 50!.

The total possible cases (all possible shuffles of 52 cards) = 52!.

Hence, the required probability is:

P(Top and Bottom are Aces) = (4C1 × 3C1 × 50!) / 52!=4/52×3/51

Option C is correct