If a coin is toss 5 times what is the probability of all heads or all tails.?
599149
The probability of getting heads on one toss is 1/2 (or 50%). To figure out the probability of getting heads on 5 consecutive tosses you just multiply the probability of a single toss 5 times:\n\n1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32\n\nThe probability of getting all tails is the same, for the same reasons: 1/32.\n\nTherefore, the probability of getting either all heads or all tails is simply 1/32 + 1/32 = 2/32 = 1/16
812841
Given 5 children there are 2^5 or 32 possible combinations of girls and boys.  Since there is only one combination with all boys that leaves 31 combinations containing at least 1 girl.  Therefore the probability is 31/32.
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
125547
You can roll 24 different sets on the two dice. (4 X 6)\nListing the 4 then the 6, you get the following totals:\n\n2:  1-1\n3:  1-2  2-1\n4:  1-3  2-2 3-1\n\nSo there are 6 ways to roll less than a 5.\n\nSo the probability is 6/24, or 1/4 which is 25%.\n\nJust to note, a 4 sided die looks like a parymid, and you actually use the number on the bottom side of the die.  I put a link to a gaming site if you want to see some odd shaped dice.
140146
Uniform distribution? So if 5, 6, 7, 8, 9, and 10 lbs of loss are all statistically equal, then since there are 6 'numbers' of weight loss, the probability would be 1/6. Think about when you flip a coin: heads or tails. Two options. Odds of it landing heads is 1/2. Same goes for tails of course.
147164
Ah, a variation on the Monty Hall problem:\n\nhttp://en.wikipedia.org/wiki/Monty_hall_problem\n\nIt is a probabilistic curiosity, this one. The highly unintuitive answer is that you actually **can** improve your chances by changing your mind.\n\nYour problem appears to only be different in that it starts with 5 boxes. The above link should give you sufficient information to solve the problem.
569858
I'll assume that he is starting with the coin.\n\nPossible scenarios: Let H be getting a head on a coin flip, T be getting a tail on a coin flip, and N be getting a nonsix on a die roll.\n\nH (Meaning he got a head on the first flip)\nTNH (Meaning he got a tail, then a nonsix, then a head)\nTNTNH\nTNTNTNH\n...\n\nSo P(H before S) = P(H or TNH or TNTNH or ...)\n= P(H) + P(TNH) + P(TNTNH) + ... since all are mutually exclusive or disjoint.\n= 1/2 + (1/2)(5/6)(1/2) + (1/2)(5/6)(1/2)(5/6)(1/2) + ...\n= 1/2(1 + (1/2)(5/6) + (1/2)^2(5/6)^2 + (1/2)^3(5/6)^3 + ...\n= 1/2((5/12)^0 + (5/12)^1 + (5/12)^2 + (5/12)^3 + ...\n= 1/2*sum(i = 0 to inf)(5/12)^i\n= 1/2(1/(1-5/12)) (since |5/12| < 1)\n= 1/2(12/7)\n= 6/7\n\nNow if he rolls the die first then possible scenarios:\n\nNH\nNTNH\nNTNTNH\n...\n\nP(NH or NTNH or NTNTNH or ...)\n= P(NH) + P(NTNH) + P(NTNTNH) + ...\n= P(N)P(H) + P(N)P(TNH) + P(N)P(TNTNH) + ...\n= P(N)(P(H) + P(TNH) + P(TNTNH) + ...)\n= P(N)(6/7)\n= (5/6)(6/7)\n= 5/7\n\nSo it depends on what was flipped first.
833257
1. There are a total of 11 coins \nA. 2/11\nB. 0\nC. 6/11\n\n2. \nA. there are 2 possibilities (5,6) and (6,5) out of 36 poswsibilities (6*6).\n1/18\n\nB. 6 possibilities: (1,6) (2,5) (3,4) (4,3) (5,2) (6,1)\n6/36 = 1/6\n\nC. There are no possibilites, the Max sum of two numbers can be 12 (two sixes)\n0\n\n\nGood luck.
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.
917829
When you are computing the probability of a given event such as 2 heads, 2 tails, or one of each, you are essentially counting the number of ways a given event can occur and dividing this number by the number of all possible outcomes.\n\nIf you toss 2 coins once ( or one coin twice) there are 4 possible outcomes:\n\nHH = both heads\nHT = 1st heads, 2nd tails\nTH = 1st tails, 2nd heads\nTT = both tails\n\nAssuming the coins are "fair" each outcome is equally likely.\n\nThere is one way in four that two heads (or 2 tails) can happen so the probability for each of these outcomes is 1/4\nBut there are two ways you can have one head and one tail and so the probability of this happening is 2/4 =1/2. The sum of all the probabilities of the distinct outcomes will always equal 1. \n1/4 +1/4 + 1/2 =1\n\nHow about if you have 3 tosses? The 8 possible outcomes are as follows:\n\nHHH \nHHT\nHTH\nTHH\nHTT\nTHT\nTTH\nTTT\nWe see there is only one way to have 3 heads (or 3 tails) and so the probability of 3 heads (or 3 tails) = 1/8\nHowever there are 3 ways you can have 2 heads and one tail, so the probability of this happening is 3/8. And there are also 3 ways you can have one head and 2 tails, so 3/8 is the probability for this too.
506414
I answered this one for you earlier...\n\nYou have X ~ Bin(6, 1/6)\n\nas you have 6 trials, with the proability of success (a four) being 1/6\n\nYou then simply need to stick this in the formula:\n\nP(X = x) = nCx × p^x × (1-p)^(n-x)\n\nIn this case you want the probability of 2 fours:\n\nP(X = 2) = 6C2 × (1/6)² × (5/6)^4 = 0.2009
631161
This is a binomial probability.\nP(success) = 1/5, P(failure) = 4/5.  There are 6 events here, so P(2 successs) = (1/5)(1/5)(4/5)(4/5)(4/5)(4/5).\nP(at least 2 successes) = P(2 successes) + P93 successes) + . . . . P(6 successes) or 1 - P(O successes) + P(1 success) + P(2 successes) which you already computed above.\nLet me know if you don't understand.
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.
1161943
Let's deal with "b" first because it makes the general solution clear.\n\nFor an unbiased coin the probability of "Heads" (H) = probability of "Tails"(T) = 1/2 = 0.5\n\nIf n = 3, then you interested in the probabilty of two people having heads and one tails; or two people having tails and one heads.\n\nThe pobability of a single event where two people have heads and one has tails = 0.5 x 0.5 x 0.5= 0.125. But this can happen 3 ways: THH, HTH, HHT and therefore the probability that 2 people have heads and one has tails = 3 x 0.125 = 0.375. Of course the probability that two will have tails and one heads will also be 0.375. So the probability that a leader would have been chosen after one iteration (vote) = 2 x 0.375 = 0.75. (Note: if 3 coins are tossed there are only 4 possibilities: all heads, all tails, two tails and one head, or two heads and one tail.)\nThe probability that a leader is not chosen the first time = 1-0.75 = 0.25. And the probablty that a leader will not be chosen after two iterations is (0.25)^2 = 0.0625. And therefore the probabilty a leader will be chosen within two iterations = 1-0.0625= 0.9375 or nearly 94%. Now we'll deal with the general case.\n\na) For arbitrary n, and a single iteration, the probability of choosing a leader = 2n(0.5)^n. And the probability of not choosing a leader is 1-2n(0.5)^n. There is a difference between the probability of choosing a leader "in" the kth iteration and "by" the kth iteration. The probability that a leader will not have been chosen after k iterations is \n(1-2n(0.5)^n)^k. And the probabilty a leader will have been chosen is simply 1 minus this value.\n\nSample calculation: assume n = 20 and k=50\n\nProb. leader not chosen = (1-40(0.5)^20)^50 = 0.9981\n\nAnd the probability a leader will have been chosen = 1-0.9981 = 0.0019 which is less than 0.2 percent or 1 in 500. This shows that as n increases, the probability that one person will have a different result than everyone else becomes very small and enormous number of iterations would be required to have a reasonable chance of choosing a leader.
304670
==========================================\nA Yahtzee is a 5 of a Kind (i.e. all the die faces are the same)\n=============================================\n\nSince there are 5 dice rolled in a yahtzee game..\n\nThe total number of possible choices \n= No. of ways in which the 5 dice can show up\n==> n = 6 * 6 * 6 * 6 * 6 \n==> n = 6^5\n\nLet "A" be the event of scoring an Yahtzee\n\nFor Event "A"\n\nNo. of favourable choices \n= No. of ways in which the 5 dice show up the same number\n==> m(a) = 6\n\n[The five dice can show either all 1's or all 2's ... all 6's]\n\nProbability of getting an yahtzee \n==> Probability of occurance of Event "A"\n==> P(A) = m(a)/n\n==> P(A) = 6/(6^5)\n==> P(A) = 1/(6^4)\n==> P(A) = 1/1296\n\n\nhttp://www.futureaccountant.com/probability/\n\n\nhttp://www.schoolingkids.com/
1076427
Probability of 10 heads, 1/2^10 = 1/1024\nProbability of 9 heads    10/2^10 = 10/1024\nProbability of 8 heads    45/2^10 = 45/1024\n\nProbablity of at least 8 heads = \nProbablity of 10 heads +  Probability of 9 heads + Probability of 8 heads  = 56/1024 = 7/128.
242656
This one is easy-assume a 50% chance each time, so its 1/2 X 1/2 X 1/2 X 1/2, so 1 chance in 16.  There is another way of figuring this, using combinations, like sn777 did, but sn777 made a logical error.  There are these possible combinations: 4 m, 3m 1F, 2m, 2F, 1m 3F, and 4F. So just looking at that, you would think its 25%. BUT, there is only one combination of 4M, while there are 4 combinations that make 3m 1F, and 6 combinations that make 2m and 2F, etc-so the total possible combinations of 4 children is 16, and 4M is only one of those.