If a coin is tossed twice, what is the probability of getting heads both times?
405425
Probability of getting one head = 1/2\nProbability of getting two heads = (1/2) * 2 = 1/4\nYou have a 25% chance of getting a head two times...\nAnd 12.5% chance of getting a head 3 times...\nAnd 6.25% chance of getting a head 4 times...\n.\n.\n.\nAnd 0.09765625% chance of getting a head 10 times!!! :)
786207
did u mean '93.5% or more'?\n\nif so, there is only 1 chance of that occurring and that is when all 8 coins are tossed and revealing heads. \n\nso that is a probability of 1/2^8 = 1/256 = 0.00390625
131088
try this site\n\nhttp://www.futureaccountant.com/probability/
716869
The probability that the first marble will be white is 3 / 5.\n\nIf we draw again without replacing the first marble, the probability that the second marble will also be white is 2 / 4.\n\n3/5 * 2/4 = 3/10.\n\nThere is a 30% chance that both marbles will be white.
284480
Each coin, of course, will come up head or tail. There are eight possible distinct outcomes:\n\nHHH\nHHT\nHTH\nHTT\nTHH\nTHT\nTTH\nTTT\n\nAs you can see, three heads and three tails come up once each, two tails and a head come up three times, and two heads and a tail come up three times.
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.
121136
need 1 more coin to show heads...\n\n4 possibilities. \n-Both coins have heads (25%)\n-1st coin has a head (25%)\n-2nd coin has a head (25%)\n- both coin has a tail (25%)\n\nadd em up.. 50%
1058790
The gambler's fallacy applies to independent random events, like coin tossing or die rolling. It's the belief that the occurrence of independent random events in the past can help you predict whether those events will occur more or less frequently than expected in the future. It's a fallacy. There's still only once chance in two that a coin will land heads-up no matter how many times it's landed tails-up recently.\n\nBlackjack is a random game, with hidden information, but which cards will be drawn in the future is influenced by which cards have been drawn in the past. In one deck, for instance, there are only four jacks, so if a jack has already been drawn, the chance of another jack appearing before the next shuffle has decreased.\n\nDon't count on being able to count cards in a casino, though. They shuffle together several decks and reshuffle drawn cards into the decks before the decks are exhausted.
765221
CASE 1:\nThe probability that 2 green balls are selected from box A:\n1/2 (chance of picking box A)\nChance of picking first ball green (3/6)\nChance of picking second ball green (2/5)\n\nSo the combined odds are: 1/2 x 3/6 x 2/5 = 1/4 x 2/5 = 1/10\n\nCASE 2:\nThe probability that 2 red balls are selected from box B:\n1/2 (chance of picking box B)\nChance of picking first ball red (4/9)\nChance of picking second ball red (3/8)\n\nSo the combined odds are: 1/2 x 4/9 x 3/8 = 12/72 = 1/6
628830
Total no of outcome : 2^11 = 2048 \n(A coin tossed 11 times)\nNo of favourable outcome : 11C4 = 330\n(Exactly 4 heads)\nSo probablility would be 330/2048 = 0.1611
479364
Yes i can say there are so many events that you can believe 50/50 chances.\n\nAs you said 50/50 is calculating for 100. means you have 2 chances.\n\n1. Toss the coin: to get head or tail you have 50/50 chance\n2. If a women is giving birth: the baby may be BOY/GIRL 50/50 Chance\n\n3. for this question : someone give the answer: which is correct/not correct   50/50 chance.\n........... \nyou can find somany events which are having 50/50 chances
600121
Your logic is flawed. Say you always pick card #2. The first time, you'll have 14.2857...% (that's the same as saying one seventh) of the chances of winning the car.\n\nThe second time, with a new set of cards, your chances that the card #2 is the winner are again 14.2857...%, and so will every new time, because your chances are always 1 out of 7. You are picking one card out of seven cards that you are given.\n\nYour reasoning would be right if you could be assured that the car card will always be a different one (that for example, it will be in the first case #4, then #6, and so on). In that case, after the 7th. guess (if not earlier), you'd get the car. But the way you explained the game, nothing prevents the car from being always for example at card #5. If that is the case, always trying card #2 will make you fail every time.
822095
a) Both: .1(.02) = .002\n\nb) Exactly one:  .02(1-.1) = .018 + .1(1-.02) = .018 + .098 = .116\n\nc) (1-.02)(1-.1) = (.98)(.9) = .882\n\nThey add up to one.  I'm not a probability person, but I think this is correct.
452389
There are 2 rules you need to know. The addition rule and the multiplication rule\n\nIf we roll a die the chance of rolling 6 is 1/6.\n\nthe chance of rolling 5 is also 1/6\n\nthe chance of rolling either a 5 or a 6 is therefore 1/6 + 1/6 = 1/3\n\nAnd the chance of rolling 4, 5 or 6 is 3/6 or 1/2.\n\nWhen the events are alternative outcomes and you want to know the chance of the outcome being one of these outcomes, add the probability of each outcome.\n\nIf you roll a die twice (or two dice at once) the probability of rolling a par of ones is 1/36.\n\nThis involves the multiplication rule. The chance of the first die showing a 1 is 1/6, and the chance of the next die showing a 1 is also 1 in 6. \n\nOnly one-sixth of the time do you get the first die to give you the 1 you want and only one-sixth of those times will the second due give you another 1. So 1/6 x 1/6 = 1/36.\n\nHowever if the question is what is the probability of just one six being thrown in two rolls\n\nthen the probability of 1 on the first throw but not on the second throw is (1/6 x 5/6) = 5/36\n\nAnd the probability of 1 on the second throw but not on the first throw is (5/6 x 1/6) = 5/36 \n\nAnd so throwing a 1 on either roll but not the other roll has a total probability of 5/36 + 5/36 = 10/36.\n\nThe slightly different question "what is the probability of rolling one or more 1s in two rolls of the dice?" is the sum of\n\nprobability of one on the first throw but not the second\n\n+ probability of one on the second throw but not the first\n\n+ the probability of a double one being thrown\n\n5/36 + 5/36 + 1/36 = 11/36\n\nSo the rule is when it is a matter of alternative outcomes to one event, add the probability of each outcome.\n\nWhen it is a matter of the outcome of two events both happening, multiply the probability of each happening to get the probability of BOTH happening.
761738
Probability(1st person picks any bill)=8/8\n\nProbability(2nd person picks the same bill)=1/8\n\nProbability(3rd person picks the same bill)=1/8\n\nThe events are independant as the outcome of one event will not affect the outcome of another. Thus allowing multiplication of probabilities.\n\nProb(All 3 draw same bill) = 8/8*1/8*1/8\n\n=1/64
630810
a) probability of rolling a 2 is 1 out 6. for 5 balanced dice\n(1/6)(1/6)(1/6)(1/6)(1/6) = (1/7776)\n\nb) isnt this the same as part a?
742241
The second problem is easier than the first problem, so I'll tackle it first.\n\nLet's imagine that the trip is 120 miles. The distance doesn't matter if the average rates are as given above.\n\nIt takes 3 hours to travel 120 miles at 40 miles per hour.\nIt takes 2 hours to travel 120 miles at 60 miles per hour.\n\nSo it took 5 hours to travel a total of 240 miles.\n\nThis is an average rate of 48 miles per hour.\n\nI'll work on the first problem and hopefully I'll get back to you.\n\nUPDATE\n\nI think I have the first problem solved as well. It involves an infinite series! Call the people involved A and B.\n\nThe probability that A wins on the first toss is 1/2.\n\nIf A doesn't win on the first toss, then the probability that B wins on the second toss is also 1/2. But this only happens in 1/2 of all games. (In the other half, A won on the first toss.)\n\nIf B doesn't win on the second toss, the the probability that A wins on the third toss is also 1/2. But this only happens in 1/4 of all games (In half the other games, A won on the first toss. In half the remaining games - 1/4 of all games, B won on the second toss.\n\nFollowing this logic, A wins:\n\n1/2 + 1/4(1/2) + 1/16(1/2) + 1/64(1/2) + ...\n\nof all games.\n\nB wins:\n\n1/2(1/2) + 1/8(1/2) + 1/32(1/2) + ...\n\nof all games.\n\nLooking again at A, we have:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...)\n\nThe term in parentheses is an infinite series having initial term 1 and ratio 1/4.\n\nThe sum of an infinite series having ratio whose absolute value is less than 1 is equal to:\n\na/(1 - r)\n\nWith a = 1, and r = 1/4, the sum = 4/3. So:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...) = 1/2 * 4/3 = 2/3\n\nLooking again at B, we have:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...)\n\nThis sum should be half of the sum of the infinite series in A.\n\nWith a = 1/2 and r = 1/4, a/(1 -r) = 2/3\n\nSo:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...) = 1/2 * 2/3 = 1/3.\n\nA should win 2/3 of the time, and B should win 1/3 of the time. There is an advantage to going first!