A family is selected at random. Find the probability that the size of the family is at least 5.?
345899
7.1 + 2.6 + 1.6 = 11.3
1143172
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
202632
The level of the student you are asking the question is to be taken into consideration for this....\n\n\ntry this site for your need...\n\n\nhttp://www.futureaccountant.com/probability/\n\n\n\nhttp://www.schoolingkids.com/
950716
The only way to fail to win at least one prize is to loose each time.\nThe probability that you will loose one time is 1 - 0.11 = 0.89.\nThe probability that you will loose 5 times is 0.89^5.\nSo the probability that you will win at least once, or fail to loose all 5 times, is 1 - 0.89^5 = 0.44
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?
1186248
In this case 'N', which is the sample size, is 120.  The probability of an event occuring is equal to the number of times it occurred in the sample divided by N, so:\n\nprobability of X=0:  20/120 = 1/6 ~ 0.1667\nprobability of X=1:  65/120 = 13/24 ~ 0.5417\nprobability of X=2:  25/120 = 5/24 ~ 0.2083\nprobability of X=3:  10/120 = 1/12 ~ 0.0833\n\nAlso the average value of X would be 1.2082 using this probability distribution through an infinite number of line checks.
323264
You are dealing with probabilities here and your odds are going to be exponential. But here is what I have for you.\n\nOdds in rolling a:\n4 = 1/12\n5 = 1/9\n6 = 5/36\n7 or 11 = 2/9\n8 = 5/36\n9 = 1/9\n10 = 1/12\n\nNow for each roll after that you would take the odds and times it to the "n" power, where "n" is the number of the roll. So to roll any of these numbers 12 times in a row would mean taking the fraction to the 12th power. Below are those answers. (Hope you understand scientific notation.)\n\n(1/12)^12 = 1.13876838470415E+55\n(1/9)^12  = 1.13771030934265E+55\n(5/36)^12 = 5.1524055910299E-11\n(2/9)^12  = 1.14868734219891E+55\n\nIn other words you would be more likely to win the lottery several times before this would happen. Hope this helps.
874576
Note: they aren't asking just for the chance that at least 2 children have blue eyes.  They are asking for the probability, given that you know at least one already does.\n\nThis is conditional probability.\n\nP(B | A) = P(A and B) / P(A)\n\nSo the probability that you have at least 2, given you have at least 1 is the probability you have at least 2 divided by the probability you have at least 1.\n\nFor that part though, you would use binomal distribution.\n\nFirst, what is the probability that at least one child has blue eyes?\nThis would be 1 minus the probability that nobody has blue eyes.  \nP ( blue >= 1 ) = 1 - (0.75)^5 \nP ( blue >= 1 ) = 0.762695312 \n\nNow the probability that at least *two* children have blue eyes:\nP( blue = 2 ) + P ( blue = 3 ) + P ( blue = 4 ) + P ( blue = 5 ).  Again you can figure the negative statement and subtract from 1.  So this would be equivalent to:\n1 - P ( blue = 0 ) - P ( blue = 1 ).\n\nThis would be:\nP ( blue >= 2 ) = 1 - (0.75)^5 - (0.25)(0.75)^4 * C(5,1)\nThe C(5,1) means the ways to choose 1 person from 5, which is just 5.\nP ( blue >= 2) = 1 - (0.75)^5 - (0.25)(0.75)^4 * 5 \nP ( blue >= 2) = 0.3671875\n\nSo putting it all together:\nP ( blue >= 2 | blue >= 1 ) = P ( blue >= 2 ) / P ( blue >= 1 )\n= 0.3671875 / 0.762695312\n= 0.481434059\n≈ 48.1%
263161
This is a good probability problem -- I LOVE probability problems.  This is really asking 4 different questions.  First, what are the chances of the first pull being black.  (9 total socks, 5 are black ...).  Then what are the chances of the second one being blue (Keep in mind that now you only have EIGHT socks in the drawer, unless you put the last one back - so 1 blue and eight total socks).  Then chances of a brown sock with only 7 socks left (3 brown, seven total socks).  \n\nWhen you get all those probabilities, in order to see what the chances of this AND this AND this happening, you have to multiply the fractions (if it was "What are the chances of this OR this happening, you'd add them).\n\nGood luck.  This stuff is FUN!  (Especially if you ever want to go to VEGAS!)
467797
The chance of not rolling a four is 5 out of 6. So five times:\n\n5/6 ^ 5 = 0.4019\n\nor about 2 in 5.
107013
Maybe there's a trick to this, but if the probability is\n5%, then if (for example) you opened 100 boxes, you would\nhave "gotten something" 5 times (ie. 5% of 100 = 5).\n\nThat seems like you'd get something, on average, once\nevery 20 boxes.
621701
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
714579
P (X>=5) \n= 1 - [  P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)] \n\nRandam sample of 100 days from the past 10 yrs... \nC(360*10, 100)
967346
This probability problem can be used using Binomial Distribution.  \n\nGiven a number of 'x' success, the probability in a binomial experiment is:\n\nP(x) = [n choose x]*p^x*q^(n-x)\n\nwhere:\n\nn choose x is a combination problem: n!/[(n-x)!x!]\np = probability of success\nq = probability of failure\nn = total selection\nx = successes\n\nIn this problem, we are looking for at least one case has a success (which is one of the highest 9 numbers).  In essence, we are solving for:\n\nP(x=1) + P(x=2) + P(x=3) + P(x=4) + P(X=5).  \n\nHowever it would be easier to find the probability of the failures and subtract from the total probability because we only have to calculate 1 probability.  This would be:\n\n1 - P(x=0) is what we are looking for.  Now all we need is to calculate the probabilty.\n\nFor P(x=0), we have the following:\n\nn = 5\nx = 0\np = (9/26) = probability of success\nq = (17/26) = probability of failure\n\nPlugging into P(x), we get:\n\nP(x=0) = (5!/5!)(9/26)^0(17/26)^5\n\nSolving, you get P(x=0) = 0.12\n\nTherefore, \n\n1 - P(x=0) is ------>\n\n1- 0.12 = 0.88\n\nBelow is a website devoted to Binomial Distribution exercises\n---------\n\nHope this helps
872621
well in my skool we take it as {<and rule>} and {<or rule>example:there is 5 white marble and 5 black marbles in a bag...u pick 2 randomly without replacing them in the bag.....wat is the probability that u get 2 white marbles==>this is a combination 2 the way it is solved is multiplying the the 2 combined probabilities\nanswer:pr:(white AND white)=(5/10)*(4/9)\n\nnow the set is this way....wat is the probability that u get 2 marbles of different colors in the first 2 picks===>set rule\n\nanswer:there is 2 combinations in which u can get 2 different colors there are {white then black} OR {black then white} u find the 2 combinations then ADD them\npr(WandB OR BandW)={(5/10)*(5/9)}+{(5/10)*(5/9)}
422184
There are 5 ways of getting a 6 (1-5, 2-4, 3-3, 4-2, 5-1) and two ways of getting 11 (6-5, 5-6), so the probability is 7/36.
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.