# Two cards are drawn from a deck of 52 cards.what is the probability that 2 cards are ACES or BLACK cards?

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In pack you will have 26 black cards and 26 red cards (which will include 2 aces). So the cards you are wanting to pull out are the 26 black cards + the 2 red aces.

So when you take out the first card the probability is 28/52.

However, once you've taken the first black/ace card, you only have 51 cards left in your pack of which only 27 now will be either black and/or ace. So the probability for the second card you take being black or ace =27/51

Therefore the probability that 2 cards are aces or blacks= 28/52*27/51= 7/13*9/17=63/221

• Total no. of outcomes=52

Favourable no. of outcomes=4(aces) + 26(black cards)=30

Probability of getting an ace or black card =30 /52

=15/26

• case1 when both are ace and not black

P1=2C2/52C2

case2 when both are black and not ace ,now out of 26 black cards u've 24 left

P2= 24C2/52C2

case3 when both r black and ace

P3= 2C2/52C2

case 3 when one is a black ace and other black and not ace

P3 =2/52*24/52

case4 when 1 is a black ace n 2nd ace not black

P4=2/52*2/52

now add all problty to get result

• aces=2/52 black=26/52=1/2

• (4/52)+(12/52)

=16/52

=4/13

• P(ace) = 13C2/52C2

P(black) = 26C2/52C2

required P = (13C2 +26C2)/52C2

= (78 + 325)/1326

= 403/1326

• 28/52 * 27/51= 28.506% for both aces and/or black cards

26/52 * 25/51= 24.50% for both black cards

4/52 * 3/51 = 0.452% both aces

• 4C2 * 13C2 / 52C2

= 6*78 / 1326

= 468 / 1326

= 234 / 663

= 81 / 221

I Think that i m right ..

• Why don't you put your question in a category where it belongs, like Mathematics?????

• (4C2+26C2)/52C2=(4/52)*(3/51)+(26/52)(25/51)=(4*3+26*25)/(52*51)=331/1326