2007-04-15 19:42:55 · 7 answers · asked by FoxBarking 3 in Science & Mathematics ➔ Mathematics
I am sorry, make that five cards.
2007-04-15 19:43:19 · update #1
Total ways to draw 5 cards = 52!/(47!5!)
Total combinations that include exactly 1 ace:
(4!/(3!1!))*(48!/(44!4!)
So ((4!/(3!1!))*(48!/(44!4!))/(52!/(47!5!)) =
= 4*48!*47!*5!/(44!*4!*52!) =
= 47*46*45*120/(6*49*50*51*52) =
= 3*23*47/(13*17*49) = about 0.3
BTW, the other answer you got had the right idea, but forgot to mutiply by 5, since the lone ace could be any one of the 5 cards.
Also, his calculation should be:
5*(1/13*48/51*47/50*46/49*45/48)
2007-04-15 19:54:09 · answer #1 · answered by blighmaster 3 · 3⤊ 0⤋
There are 4 Ace cards in deck of 52, hence 48 non-Ace cards in the deck.
You are going to draw 5 cards in total but only 1 being an Ace. Hence here's the number of choices for each of these 5 cards:
Card1: 4 (being an ace)
Card2: 48 (being the 1st non-ace)
Card3: 47 (being the 2nd non-ace)
Card4: 46 (being the 3rd non-ace)
Card5: 45 (being the 4th non-ace)
Mathematically,
No of ways of getting exactly 1 Ace in 5 cards from a 52-cards deck
= (4C1)*(48C4) = 778320
Total number of ways of getting 5 any cards
= (52C5) = 2598960
Probability of getting exactly 1 Ace in 5 cards from a 52-cards deck
= (No of ways of getting exactly 1 Ace in 5 cards from a 52-cards deck) / (Total number of ways of getting 5 any cards)
= 778320 / 2598960
= 0.29947363560808938960199464401145
2007-04-16 03:04:15 · answer #2 · answered by QiQi 3 · 0⤊ 0⤋
the probability that u draw an ace is 4/52 ,for 5 cards drawn
maybe the first one will be an ace OR the second one OR the third....
So Probability= 4/52 +4/52+.....=20/52=0.3846
2007-04-16 03:24:31 · answer #3 · answered by Anonymous · 0⤊ 2⤋
All of these answers on here are wrong...You have to assume random sampling in all drawings which mean you have to put the card back you just drew...So, this one is easy.. and I have taught statistics so trust me..
1/52 + 1/52 + 1/52 + 1/52 + 1/52 = 5/52 or .0961 or 9.61%
2007-04-16 03:16:20 · answer #4 · answered by okstatecowboy 4 · 0⤊ 2⤋
Assuming card is replaced B4 drawing other
(4C1)* ((4/52) ^1) * ( (1-4/52) ^3 ) =0.242
Repeated trial
Am i right??
Tell me
Thanks
2007-04-16 02:56:48 · answer #5 · answered by Muhammad 2 · 0⤊ 0⤋
You have to draw one ace (1/13) and four "not aces" (12/13 each). In probability, AND means to multiply.
1/13*12/13*12/13*12/13*12/13
=20,736/371,293
=0.0558 or 5.58%
2007-04-16 02:51:02 · answer #6 · answered by ecolink 7 · 0⤊ 2⤋
.3846 to 1
2007-04-16 03:09:32 · answer #7 · answered by lahomaokie 2 · 0⤊ 0⤋