This is a Statistics (Probability) question. I know there are six different ways the cards can come up. I just need to find the probability of each of the six events and then find the total probability of all six events.
126071
if the order you draw them is important its a permutation.\n\nthe answer is basically; a 1 in 55 chance for the first card, a 1 in 54 chance for the second card (you've taken the first card out) and a 1 in 53 chance of getting the final card. all three events are independant of eachother. so the answer is 55 x 54 x 53 = 157410\nso 1 in 157410 chance.\n\nif the order you draw them is unimportant them its a combination.\nfor the first card there is a 3 in 55 chance\nthe second card has a 2 in 54 chance\nand the final card has a 1 in 53 chance\nthe answer is: (3/55)  x (2/54) x (1/53) = 1 in 8745
178452
try this site for help on probability\n\nhttp://www.futureaccountant.com/probability/
656018
If you're accepting the thought that every hand is random and that the probability of winning any hand is 1/2, then you're correct.\n\nThis would be a binomial distribution.  The probability of winning at least one hand = 1 - (the probability of losing all four).\n1 - (1/2)^4 = 1 - 1/16 = 15/16.\n\nThe only fallacy to the actual process is that the probability of winning subsequent hands after the first isn't going to be exactly 1/2, as the cards that were used are gone from the deck and change the odds for the rest of the deck.  But if you're looking for a good ballpark figure, you're on the right track.
120375
(5/15)*(6/14)*(5/13)*(4/12) = 600/32760 = 5/273. This is correct because the order you picked cards matters.
54871
each child can be a male or a female \ntherefore the total no of events possible is 2^6=64\n\nthe question says that you should have atleast 3 girls \ni.e there can be three, four ,five or all six girls .\n\nthe number of ways of choosing 3,4,5,6 girls from 6 children is the sum\n6C3 + 6C4 + 6C5 + 6C6 (here C stands for combinations )\n\ni,e\n\n(6*5*4)/(3*2*1) + (6*5*4*3)/(4*3*2*1) + (6*5*4*3*2)/(5*4*3*2*1) + 1\nthis reduces to \n20+15+6+1\n=42\n\ntherefore the probability is \n42/64
670801
The probability is 11/1105.\nYou mulitply all of the fractions- 12/52, 11/51, 10/50 and than simplify.\nBy the way, a standard deck has 52 cards, not 53 cards.
279773
Make a tree that shows all possibilities you have.\nOn the first level there is the choice of the box with 0.5 probability each. On the next level, box A, you have two choices again, with the prob. is 3/7 to find a red ball, 4/7 to find a white ball, etc.\n\nWhen you have finished to tree, use it to make a list and\ncalculate the prob. for picking red-red, red-white, etc.
636892
There is definitely something wrong with the wording of the puzzle, because you can just turn over the [A] card. Anyways, I doubt your problem is that easy. I'm guess that you're talking about the reverse of both Aces.\n\nSince each card card will have only a number and a letter, there are only 4 possibilities:\n\n1)\nFront: [A] [K] [2] [7]\nBack: [2] [7] [A] [K]\n\n2)\nFront: [A] [K] [2] [7]\nBack: [7] [2] [A] [K]\n\n3)\nFront: [A] [K] [2] [7]\nBack: [2] [7] [K] [A]\n\n4)\nFront: [A] [K] [2] [7]\nBack: [7] [2] [K] [A]\n\nSince there are actually 2 Aces, you could play a trick question and flip over the other ace as well. So even if there is a 2 behind the [A], there might be another Ace behind the [2]. The easy way is to flip over the [A] and the [2].\n\nHowever, you can do it 3 other ways. By flipping over exactly 1 face card (A or K) and 1 numbered card (2 or 7), you can determine exactly what's behind all four cards.\n\nFor example, if you flip over the [7] and the [A]:\nIf a 7 is behind the [A], then the 2 is behind the [K], and vice versa. If the A is behind the [7], then the K is behind the [2], and vice versa.\n\nOnce the combination is found, you will know if either 2 is paired with either Ace. (If they are randomly sorted, the probability will be 3 out of 4).
508814
If the die were perfectly balanced, six-sided cube, and had the each of the numbers 1 through 6 on it, then you don't need the tally.\n\nThe odds of any number coming up are 1 in 6.\n\nThe odds of any number coming up five times in a row are:\n1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 6= 6/6^5  = 6/7776, or,  1 in 1296\n\nHope this helps!
367257
a) There are 16 combinations and 4 of them include exactly 3 heads (HHHT, HHTH, HTHH, THHH).  So the probability is 4/16 or 1/4.\n\nb) This question results in the same probability (4/16 = 1/4), except it asks for it as 'odds' which is the probability of success divided by the probability of failure.  That is 1 to 3 odds.\n\nc) The odds against heads are just the reverse.  Probability of failure divided by probability of success.  That is 3 to 1 odds against.
178489
In game one between A and B, the outcomes are:  A wins, B wins, or they draw, equaling 3 outcomes.\n\nIn 14 games, there are 3^14 outcomes.
45354
If you have two events "A" and "B"\n\nThe probability that\n\na) Both would occur = P(A intersection B)\nb) Only one would occur = P (A intersection B') + P (B intersection A')\nc) At least one of them would occur = P (A union B)\nd) None of them would occur = P (A' intersection B')\n\nTry this site for a good explanation on probability\n\n\nhttp://www.futureaccountant.com/probability/\nhttp://www.schoolingkids.com/
96549
Remember that AND implies multiplication of probabilities while OR implies addition. This can be derived from the fundamental law of counting.\nConsider question (a). In conditional probability, the probability that two independent events A and B occured is the product of the probability that A occured and the probability that B occured given A's occurrence. Consider this definition logically (Use Venn diagrams). In symbols P(A AND B) = P(A)*P(B|A).\nThe statement (a) is saying that you know you drew a red ball, but you want to know the probability that you drew it from box A. Thus, we want to know the conditional probability P(I draw a ball from box A | I draw a red ball) = P(I draw a ball from box A AND it is red) / P( I draw a red ball ) = [(1/3)*(3/5)] / [(1/3)*(3/5 + 1/3 + 4/7)] = (1/5) * (3*3*5*7)/(3*3*7 + 5*7 + 3*4*5) = 63/158.\nNote that this is not the same as the probability that you draw a red ball from box A, which is just (1/3)*(3/5), because the implicit condition of that statement is "if I draw a ball" whose probability is just 1.\nThe other questions are similar.
192530
First choose which number it is.  There are 6 numbers, so there are C(6,1) = 6 choices.\nThe probability of that number is 1/6\nSince there are three\n(1/6)(1/6)(1/6) = 1/216\nNow multiply this times 6\n6/216 = 1/36
427357
Theoretical probability is finding the probability of events that come from an equiprobable sample space or, in other words, a sample space of known equally likely outcomes. \nFor example, \nSay you roll a dice.\nThe probability of rolling a 1 is the same if you roll a 2, 3, 4, 5, or 6.\nThat is, each number is equally likely.\nSince there are 6 numbers in the sample space, the probability of each event is 1/6\n\nAnother example:  Say you flip a fair coin.\nYou have two possiblities.  It could land heads, or it could land tails.  Each event is equally likely, so the probability of landing heads is the same as the probability of landing tails (which is 1/2)
240516
Actually, the above answers are incorrect.\n\nWe have 5 cards and 4 suits.\n\nLet's, say we draw the cards in the following order:\nAce, 2,3,4,5\n\nThere are 4 suits. So there 4 ways we could draw an Ace first; 4 ways we could draw a 2, 2nd, etc. And so there are 4x4x4x4x4 = 1024 ways of drawing the cards in the above order. However the cards can be drawn in any order, so we have to multipy 1024 by the number of ways the five cards  could be ordered when they are drawn. The number of ways the 5 cards could be ordered is 5 factorial (i.e 5!) = 5x4x3x2x1=120. So the number of ways of drawing a hand consisting of the 5 specified cards, in any suit and in any order  =  1024x120 = 122,880
242624
1.  I'm assuming you mean "exactly one even card."\nP(1st card is even and 2nd card is odd and die is even) +\nP(1st card is odd and 2nd card is even and die is even)\n= (20/40)(20/39)(1/2) + (20/40)(20/39)(1/2) = 0.256\n\n2.  P(1st card is even and 2nd card is even and die is even)\n=  (20/40)(19/39)(1/2) = 0.123\n\n3.  Again, I'll assume you mean "exactly one card of 3 is drawn."\nP(1st card is 3 and 2nd card is not 3 and die is less than 3) + P(1st card is not 3 and 2nd card is 3 and die is less than 3)\n= (4/40)(36/39)(2/6) + (36/40)(4/39)(2/6) = 0.062\n\n4.  P(115 or 124 or 133 or \n142 or 151 or 214 or \n223 or 232 or 241 or \n313 or 322 or 331 or \n412 or 421 or 511) = \n(4/40)(3/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + \n(4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + \n(4/40)(3/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + \n(4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(3/39)(1/6) + \n(4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(4/39)(1/6)\n= 0.024\n\n5. P(111 or 112 or 121 or 211) =\n(4/40)(3/39)(1/6) + (4/40)(3/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(4/39)(1/6)\n= 0.0060