Two cards are drawn from the top of a well-shuffled deck. What is the probability that they are both black aces?

There are

52

cards in a deck, and there are

4

aces in a deck, so the probability of drawing an ace is

4

52

If you do not replace the card, there are only

51

cards left and only

3

aces left, so the probability of now drawing an ace is

3

52

# Multiply these probabilities:

4

52

x

3

51

12

2652

→

0.00452

but the answer is wrong.because in question it is clearly mentioned that i have to pick up the black ace.

and please tell me what is the wrong in my approach-

A =that the cards must be black

so p(A)=1/2

B=probability of getting 2 aces.

p(B/A)=probability of getting(2 aces which should be black)=2/26*1/25
p(A and B)=p(A) p(B/A)
=1/22/26*1/25

=1/650

you have picked all.so can you correct it now?

without replacement:

There are 52 cards in a deck, and there are 2 aces in a deck wit black color, so the probability of drawing an ace is 2/52.

If you do not replace the card, there are only 51 cards left and only 1 black

ace left, so the probability of now drawing an ace is 1/52.

Multily both these probablities : (2/52) * (1/52) = 0.0007396

with replacement:(2/52) * (2/52) =0.00147

[quote=“iftekar-patel-f1e6bf65, post:7, topic:3032, full:true”]

without replacement:

There are 52 cards in a deck, and there are 2 aces in a deck wit black color, so the probability of drawing an ace is 2/52.

If you do not replace the card, there are only 51 cards left and only 1 black

ace left, so the probability of now drawing an ace is 1/52.

Multily both these probablities : (2/52) * (1/51) = 0.0007541

with replacement:(2/52) * (2/52) =0.00147

but what is the wrong in above approach-Probability question.