How do you solve this probability question?
968185
First the probability that a family of 4 is all girls is:\n1/16 (and the chance that they are not is 15/16).\n\nRather than figuring the chance of at least one family having this arrangement, figure the probability that *none* of the families do.  Then subtract from 1.\n\nThe answer will be 1 - (15/16)^10\nor about 47.6%\n\nIt's not a simple multiplication by n... if it were, then with 16 families you should have a 100% chance.  And with 32 families you would have a 200% chance.  That logic would be correct if you couldn't repeat a combination, but you can.  Just because family A has four boys, for example, doesn't mean family B can't also have four boys.\n\nLabel the 10 families as A, B, C, D, ... , I, J.  The probability that A has four girls is 1/16.  The probability that B has four girls is 1/16, etc.\n\nThe chance that *none* of them have four girls is 15/16 x 15/16 x 15/16 x ... x 15/16.  This is (15/16)^10 or about 52.4%.  Therefore the chance that at least one of them has four girls is 100 - 52.4 or 47.6%
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%
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.
1092672
Its 1:220\n\n(4/52)(3/51) = 1/221 which is the probability\nThus the odds are 1/220 or 1:220
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1.Probability of getting a number from 1-4 on throwing the dice\n         = 4/6\n\n2.Probability of a correct throw = 1/4\n\n3.Total number of questions = 160\n\nYour Answer = (4/6 * 1/4) raised to the power 160
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1) Find the probability of getting a red ball when a bag contains 2 red balls and one blue ball.\n\n2) There is a red ball, a blue ball and a green ball in a bag. Find the probability of getting either a red ball or a blue ball.\n\n3) There are 10 yellow chips and 20 purple chips in a sack. When one chip is selected at random, find the probability of NOT getting yellow chips.
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In the experiment of tossing a coin ten times.\nTotal number of possible choices = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 \n\n      = 1024 \n => n = 1024\n\nLet A be the event of getting number of heads is double the number of tails.\n\nFor event A\n------------\nNumber of favourable choices  =  Number of chances of getting heads is double the number of tailes.\n\n                  = No chances for occurance of event A\n                  = 0\n           =>  m  = 0\n\nProbablity of occurance of an event A =  Number of favourable choices / total number of possible choices\n     => P(A) =  m/n\n     => P(A) =  0/1024\n     => P(A) =  0\nTherefore the probability is zero.\n\n\n\nFor all about probabilty:\n\nhttp://www.futureaccountant.com
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.
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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.
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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
554382
Matthew rolled four dice. What is the probability of the total of the spots equaling 14? The answer is 1/6.\n\nJenna decided to repeat Matthew's experiment, except she rolled six dice. What is the probability of the total of the spots equaling 21? The answer is 1/6.\n\nLarry decided to repeat Jenna's experiment, but with ten dice this time. What is the probability of the total of the spots equaling 35? The answer is 1/6.\n\nGive a formula as to show when 2n dice are rolled, which total of dice will have a probability of 1/6 of being rolled.
249369
Is the term you are looking for "probability"?\n\nI hope this helps.
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Suppose the people are p1, p2, p3, p4, p5, p6, p7.\nThe probability that p1 was born on Monday and nobody else was is:\n1/7 * (6/7)^6 where 1/7 is the probability a person is born on Monday and 6/7 is the probability someone isn't born on Monday.\nThe same is true for the probability that p2 is born on Monday and nobody else is, etc.\nSo the total is\n7 * (1/7)*(6/7)^6\n= (6/7)^6\n= 46656/117649\n= .3965694565
239098
Easiest way to solve it:\nthe probability that the first one FAILS is \n1.00 - 0.95 = .05 (5 percent of the time)\nprob that the second FAILS is \n1.00 - 0.92 = 0.08\nThey are what is called "independent events" because they are on different circuits.  So, the probability of two independent events is the product of the chance of each happening by itself.  \n0.05 * 0.08 = the probability that BOTH fail.\nI'll let you do the math and finish the problem.
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Here's the formula:\nThe probability that all birthdays are different is\nP(365,r)/365^r, where r is the number of people and P stands for permutation.\nThus, the probability that at least two are the same is\n1 - P(365,r)/365^r\nr = 20\n1 - P(365,20)/365^20\n= 1 - .588561616419\n= .411438383581\nSo there is a .4114 chance that at least 2 people in a group of 20 people will have the same birthday.\nThe birthday problem is pretty interesting. If there are 23 people, the probability that at least two have the same birthday is .507. If there are 50 people, it is .970.
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Probability\n= P(boy and anyday) and P(boy and that same day)\n= (1/2)(1) * (1/2)(1/365)\n= (1/2)*(1/730)\n= 1/1460
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try\nhttp://www.peterwebb.co.uk/probability.htm