What are the formulas for using probability with only, both, at least one, and neither?
114803
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/
92016
When dealing with the occurrence of more than one event, it is important to be able to quickly determine how many possible outcomes exist.\n\nFor example, if ice cream sundaes come in 5 flavors with 4 possible toppings, how many different sundaes can be made with one flavor of ice cream and one topping?  \n\nRather than list the entire sample space with all possible combinations of ice cream and toppings, we may simply multiply 5 • 4 = 20 possible sundaes.  This simple multiplication  process is known as the Counting Principle.
1011849
a) probability of finding exacttly 2 working bulbs in first three =  3*( 9/12*8/12*3/10) = 27/55.\n\n\nb)\nprobability of not finding even one = 3/12 * 2/11 * 1/10 = 1/220\nprobability of finding exactly 1 = 3 * ( 9/12 * 3/11*2/10)  = 9/220 \n \nthus required probability = 10/220 = 1/22.
885691
Part (a) of problem 1 is pretty easy.  There are 10 marbles total, and 6 of them are red.  On your first draw, you have a 6/10, or 0.6 (60%) probability of drawing a red sock.  In other words, there are 6 ways of drawing a red sock out of a total of 10 possibilities.\n\nFor part (b), you need to know that the probability of two independent events happening is given by the product of the probabilities of each of the individual events.  In this case, the two independent events are that the first marble drawn is red AND the that the second marble is also red.  We already calculated the probability for drawing a red marble on the first draw in part (a).  After the first draw, the box contains 9 total marbles, 5 of which are red, so the probability that the second marble is red is 5/9.  The probability that both are red is given by the product, 6/10 * 5/9 = 30/90 = 0.3 (30%).\n\nThe reasoning for part (b) is the same as for part (a).  The probability that the first marble drawn is white is 4/10, and the probability that the second marble drawn is also white is 3/9.  The probability that both are white is given by the product, 4/10 * 3/9 = 12/90 = 0.133... (13.333...%).\n\nFor part c, there are two ways one can get a red and a white marble: either the first marble is red and the second white, OR the first marble is white and the second red.  When there is more than one way to achieve the desired result, the probability of getting that result is given by the SUM of the probabilities of each of the different ways to achieve that result.  In this case, the probability of drawing red-white is 6/10 * 4/9 = 24/90, while the probability of drawing white-red is 4/10 * 6/90 = 24/90.  The total probability of getting a red and a white marble is then 24/90 + 24/90 = 48/90 = 0.53333....\n\nUsing the same reasoning, you should now be able to solve parts (a) and (b) of the second problem.  You should find for part (a) that the probability is equal to 0.42424242...., and that the probability for part (b) is 0.09090909....\n\nFor part (c), you would use a similar logic as used for part (c) of the first problem.  The probability of getting a matched pair (either both white or both blue) is given by the sum of the probabilities found in parts (a) and (b).  You should find that the probability is 17/33 = 0.51515151....\n\nFor the third problem, you need to know that the expected "payoff" of a game is given by the sum of the products of the prize value times the probability of winning that prize.  In this case, the probabilities of winning the various prizes are:\n\n$100 : 1/500\n$50 : 3/500\n$25 : 5/500\n\nSo the expected payoff (i.e., the "mean" or "average" winnings for a single ticket) is given by:\n\nP = $100*(1/500) + $50*(3/500) + $25*(5/500)\nP = $100/500 + $150/500 + $125/500\nP = $375/500\nP = $0.75
590394
In 36 possibles outcomes\nprobability of doubles\n1-1\n2-2\n3-3\n4-4\n5-5\n6-6\n6 in 36 or 1 in 6 or 16,666%\nOf the sum being 4\n1 +3\n2+2\n3+1\n3 in 36 or 1 in 12 or 8,333%\nIf it is combined probability you can sum the two\n16,66+8,33=25%
998556
Find the probability of the complement occurring and then subtract that from 1.\nThe complement would be that an odd occurs 0 or 1 times.\nP(odd) = .5 and P(even) = .5\nP(0 odds) = C(10,0)(.5)^0 (.5)^10 = .0009765625\nP(1 odd) = C(10,1) (.5)^1 (.5)^9 = .009765625\n.0009765625 + .009765625 = .0107421875\n1 - .0107421875 = .9892578125
257322
It's very simple. Every taster has to choose one out of 5 cups, so on every round the probability of a single taster choosing A is 1/5.\nSince the event is repeated 4 times independently, the overal probability will be 1/5 of 1/5 of 1/5 of 1/5.\nThat is, (1/5)^4 = 1/625.
850369
The answer to this question is (b) 2/7.\n\nWhen removing two dogs there are four possible scenarios:\n\n(1) Remove poodle, then remove terrier.\n(2) Remove poodle, then remove another poodle.\n(3) Remove terrier, then remove another terrier.\n(4) Remove terrier, then remove poodle.\n\nIn each case, take the probability of the first even happening and multiply it by the probability of the second even happening. For instance:\n\nProbability that a poodle is first selected = 5(# of poodles originally) / 7(total # of dogs).\nProbability that a terrier is then selected = 2(# of terriers left) / 6(total # of dogs after a poodle is removed).\nMultiply 5/7 and 2/6 to get 10/42.\n\nRepeat this with each of the remaining three cases to get\n(1) 10/42\n(2) 20/42\n(3) 2/42\n(4) 10/42\n\n(Note that these add up to 1, which means that we have all of the possible cases)\n\nNow, just add the probabilities of the two cases with a terrier being selected second to get:\n\n10/42(first case) + 2/42(third case) = 12/42\n\nReduce 12/42 down to 2/7 to get your answer.\n\nI hope this helps, good luck with future probability problems!
1015803
The basic equation of probability is P(A)= Limit(NA/N) as N approaches infinity. Where \nA is an event\nNA is the number of occurrence of event A over a population sample N \nN is the to total number of samples (N does not have to infinite but sufficiently large)\n\nFor example if a pole is taken  of a population size of 1000 people by asking them the same question and 400 of them answered with “No” then P(“Yes”)=600/1000 and P(“No”)=400/1000 \n\nAs soon as you asked a question and included “A or B” statement you have opened … No that is okay, just more info.\n\nWhen A and B are two independent events then the probability that events A and B both occur is: P(A and B) = P(A) x P(B).  The outcome of one toss of a coin is independent from your next toss. \n\nWhen A and B are mutually exclusive, then the probability of A or B is simply: \nP(A or B) = P(A) + P(B).\nThe outcome of one toss of a coin will yeld either heads (H) or tail (T) since both H and T are mutually exclusive.\n\nI hope that helps
911469
I'll do some for you to show how they are done. In all cases probability P = #number of good throws/ # of all throws.\nAlso 1-P = #number of bad throws/ # of all throws\nOften you can arrive at a solution by considering the dices one by one.\n1) First dice can be everyting, second dice can be one of the five numbers not yet thrown, third dice can be one of the four numbers not yet thrown, etc. This leads to P = (6/6) * (5/6) * ... *(1/6) =720/46656 = 0.0154\n2) Divide the dice in three pairs, named A,B and C. The probability to throw a pair of two's with pair A is 1/36. Idem for B and C. But we have 6 posibilities to arrange the pairs ABC, ACB, ... And we have 6x5x4x3x2x1 possibilities to arrange the dice into the three pairs. Therefore P = 6 x 720 x (1/36)^3 = 0.0926\n3) First dice 1, second dice 1, third dice 1 gives a probability of (1/6)^3 = 1/216. But there are 20 possibilities to arrange the dice.\nP=20/216 = 0.0926
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.
1136735
P(Last is W) = P{(2nd is W and 1st is W) or (2nd is B and 1st is W)}\n= P(2nd is W and 1st is W) + P(2nd is B and 1st is W) (since these are mutually exclusive events)\n= P(2nd is W)P(1st is W|2nd is W) + P(2nd is B)P(1st is W|2nd is B) (both use the general multiplication rule)\n= (6/10)(7/11) + (4/10)(6/11)\n= (42+24)/110\n= 66/110\n= 0.6\n\nWhich is the probability of getting a white from either box, which I think makes sense, since both have the same proportion of whites.
65826
one sided only has the upper limit and anything below/lower limit and anything above\n\n2 sided gives both the upper and lower range for the value to fall in between
97799
i don't believe the person above me got the second part right.  he is right about the beginning - the first probability is 2/5:\n\npossible numbers to draw: 2 4 6 11 13 15\n\nso you can choose any of these six out of the fifteen given (6/15 = 2/5).  the next number you draw must be also a "winning" number.  BUT, you have to have an even number less than 7 AND an odd number greater than 10.  this gives us two scenarios:\n\nIF the first number picked was even and <7, then the possible numbers you can pick from are ONLY: 11 13 15, since picking another even number <7 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\n\nIF the first number picked was odd and >10, then the possible numbers you can pick from are ONLY: 2 4 6, since picking another odd number >10 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\nSO, finally, suppose you go with the first scenario.  the probability would be: (6/15)*(3/14) = 3/35  going with the second scenario will give you the same thing, 3/35.  thus adding these two, since you can have either scenario occur and still meet your requirements, gives you: 3/35+3/35 = 6/35.\n\nAnd that would be the odds: 6/35.
1097980
0.8037 or 80.37%\n\nThis is why:\n\nP(E)=1-P(not E); so the probability of at least 2 people having the same number is 1- the probability that each person has a unique number.\n\nThe probability that each person has a unique number:\n\nFirst person has 100 choices out of 100 numbers, so 100/100.\nSecond person can't choose the first person's number, so only 99 choices out of 100, so 99/100;\nThird only 98/100;\n. . .\n18th person, has (100-17)/100=84/100\n\nTherefore the probability that none of these people have the same number is (100/100)•(99/100)•(98/100)•(97/100)•. . .•(84/100) ≈19.63%\n\nSo the probability that at least two people have the same number is 100%-19.63%=80.37%.
564027
At least two of them, or exactly 2 of them?\nOf couse, assuming they choose just one recipe for "supper"\n\nProb. that at least 2 of them choose the same recipe.\n\n1 chooses a recipe:\nprob 2 picks diff. recipe = 7/8\nprob 3 choose diff than 1 and 2 = 6/8\nprob 4 diff than all above = 5/8\nprob 5 diff than all above = 4/8\n\nprob they all choose diff = 7/8 x 6/8 x 5/8 x 4/8 = 0.205\nprob at least 2 choose same = 1 - .205 = 0.795 or about 80%\n\nprob exactly 2.\n\n1 chooses recipe\n2 chooses same = 1/8\n3 chooses diff = 7/8\n4 chooses another diff = 6/8\n5 chooses another diff = 5/8\n\nprob 1 and 2 choose same (rest diff) = 0.0513\nhowever 2 and 3 (or any pair) could choose same (rest diif)\nnumber of possible pairs from 5 is (5)(4)/2 = 10\nSo probability that exactly two are same = 10 x .0513 = 0.513 or about 51%. \n\nnote: prob exactly 2 (51%) is less than at least 2 (80%)
937552
You can do this two ways...\n\n1. Find P(x=1) + P(x=2) + P(x=3).\n\n2. An easier way: find P(x >= 1) by taking P(x = 0) and subtracting that from 1.\n\nEither way, from the table, you should get the same thing.\n\nedit: That rule of thumb will work for the binomial distribution because it is discrete.  It would not for, say, continuous distributions.  A better rule of thumb might be this...\n\nP(X >= x) = 1 - P(X < x).\n\nThis will always work.  Hopefully you understand the reason behind it; the probability of an event is equal to one minus the probability of the opposite of that event.