Question

Two cards are drawn from a regular deck of 52 cards, without replacement. what is the probability that the first card is an ace of clubs and the second is black?

Answer 1

Type this into your calculator (1/52)×(25/51) then convert to a percentage or fraction of leave as a decimal if that's what it asks for.

There is only 1 ace of clubs and it is black, so the first probability is 1/52. One ace of clubs out of 52 total cards.

Next, you keep the first pick and want a black card second. You subtract the one card you took from the total deck, leaving you with 51. (btw With replacement, you would simply keep the total as 52) There are 26 red cards and 26 black cards in a deck. It would be 26/52, but since the ace of clubs is black, it is 25/51.

Finally, you multiply both probabilities to get the overall probability (lol weird way of saying that, but I hope you get what I mean).

Wanted/Total

Answer 2

The probability  when first card drawn is an ace of clubs and the second drawn is black is $\frac{25}{2652}$

Given

Deck of 52 cards. Two cards are drawn with out replacement .

There are 4 ace in a deck

1 ace of club in 4 ace of deck .

There are 13 spade and 13 clubs in a deck .

So totally there are 26 black cards

Probability of selecting first card = $\frac{ace \; of c\; lub}{total \; cards} =\frac{1}{52}$

first card is selected . so 51 cards left in a deck

one Ace black club is already drawn. so 25 black cards left.

Probability of selecting second card =$\frac{26}{51}$

Probability (ace of club and second is black)=$\frac{1}{52} \cdot \frac{25}{51}=\frac{25}{2652}$

Learn more :   brainly.com/question/16928368