Three fair coins are tossed.  Find the probability of getting the same thing on all three coins.?
349201
Write out the sample space:\nHHH, HHT, HTH, THH, HTT, THT, TTH, TTT.\n\nThere are 8 different possibilities - 2^3.\n\nTwo of the outcomes: HHH and TTT are all the same -\nProbability is 2/8 = 1/4 = .25.
395430
Assuming that the chance of having a girl is 50%, the chance would be 3/8.  This answer can be most simply reached a couple different ways.  First, you could list the possibilities:\n\nBBB\nBBG\nBGB\nGBB\nGGB\nGBG\nBGG\nGGG\n\n3 of the 8 possibilities fit the condition (exactly one girl), so the probability is 3/8.\n\nAnother way, better for larger numbers where the above method is impractical, is to use permutations.  The above works fine for this problem, though.
885606
I think this is only hard if you make it hard, but it's quite easy if you look at it in the right way. Since the order the cards are dealt in is arbitrary, you can assume that that the first card that was dealt was one of the two identical ones, and we can call the person who received it 'Alan' (note we are naming him AFTER he got the card, so we're not finding the probability of it going to a particular player). There are now 51 equally likely 'slots' that the other identical card can go in, and 12 of them are 'slots' that would give Alan the other identical card. So the probability is 12/51 .
938617
You can read off the outcomes from Pascal's Triangle. For 5 tosses consult the fifth row\n\n........1  1\n.......1  2  1\n......1  3  3  1\n.....1  4  6  4  1\n....1 5 10 10 5 1\n\nThere is \n1 chance in 32 of 5 heads and 0 tails\n5 chances in 32 of 4 heads and 1 tail\n10 chances in 32 of 3 heads and 2 tails\n10 chances in 32 of 2 heads and 3 tails\n5 chances in 32 of 1 head and 4 tails\n1 chance in 32 of 0 heads and 5 tails
90357
F- even and odds.  There are 3 even sides and 3 odd sides. So the likelihood is the same for both.
95999
Don't you think you could have written this more clearly? \n\nIf there is a uniform distribution of the weight w being between 5 and 10, then each of these probabilities is equal:\n\nP5_6=P{5<=w<6}\nP6_7=P{6<= w<7}\nP7_8=P{7<=w<8}\nP8_9=P{8<=w<9}\nP9_10=P{9<=w<10} \n\nThe sum of these is 1, so \n\n1=P5_6+P6_7+P7_8+P8_9+P9_10, and\nP5_6=P6_7=P7_8-P8_9=P9_10\n\nso 1=5xP9_10, so P9_10=0.2\n\nThe probability is 0.2.
332812
A fair dice is one in which the probability of getting one side is 1/6.\n\nSum of 6 or 11 can be obtained in the following ways: \n(1,5) (2,4) (3,3) (5,1) (4,2) (5,6) (6,5)\n\nThere are a total of seven elements here.\nTotal possible elements = (6 X 6) = 36\n\nHence probability = 7/36
941928
if you are lucky you will pick 8 balls of the same colour. \nand the question is how lucky are you??
727752
Number of all permutations = 8!\n\nNumber of all permutations with 5 boys first (then 3 girls) = 5!3!\nNumber of all permutations with 5 boys last (3 girls first ) = 5!3!\n\nSo the probability in question is 2*5!3!/8! = 1/28
577657
since it is "at least", use 1-(probability of no aces or no kings), so:\n\n1-((44/52)*(43/51))\n\nfor the second problem, I would use the idea of (probability of no orange + probability of one orange), so:\n\nprob of no orange:\n((5/8)*(4/7)*(3/6)*(2/5))\n\nprob of one orange, assuming you've learned combinations and permutations:\n(3C1*5C3)/8C4\n\nadd the two prob from above.\n\nhope this helps...probabilty can be quite ambiguous.
629548
3. Each get one third of the prize. Not fair
600596
There are 6 possible outcomes for the roll. There is only one outcome when the number is 4. Therfore the chances of getting a 4 are 1/6.
733920
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.
496704
Assuming that the 1, 2 and 3 are supposed to be consecutive:\n\nFirstly, for three throws of the die, the probability of throwing 1, 2 and 3 in order is (1/6)^3 = 1/216\n\nIf you throw the die 10 times then there are 8 independent chances of throwing 1, 2, 3. That's because if you have not thrown 1, 2, 3 by the time you get to the 9th throw, you cannot do it because you only have two throws left. \n\nTherefore the chance of throwing 1, 2, 3 with 10 throws of a die are (10 - 2) * (1/216) = 8/216 which is equivalent to 1/27.\n\nIf you prefer to express that as a decimal (not recommended!) then the answer is 0.0370 to 3 s f
192695
You have the following lists:\nB B B\nB B G\nB G B\nG B B\nB G G\nG B G\nG G B\nG G G\n\nSo you have 8 variations and of those 3 are 2 boys 1 girl so the probability is 3/8.
138694
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
1179837
a) Try counting the possibilities.\n     There is only one way to roll a 12...(6, 6)\n     and there are 36 different possible rolls.\n     so the chances are 1/36\n\nb) There are 2 possibilities for this one (3, 6) and (6, 3)\n     so it's 2/36 or 1/18\n\nThe difference between the book the you is notation.\n1:35 means that for every one success there are 35 failures.\n1/36 means that there is one success for every 36 tries.