4,879 views
23 votes
23 votes

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

  1. $\frac{4}{52} \times \frac{4}{52}$

  2. $\frac{4}{52} \times \frac{3}{52}$

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

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

4 Answers

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

Correct Answer: $C$
edited by
22 votes
22 votes
$E_1$ : 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: $C$
edited by
1 votes
1 votes
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
1 votes
1 votes

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:

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