Probability-What is the probability that the shipment is rejected? Is not rejected?
960827
Both answers above are wrong.  It's obvious that the quality control will inspected the random samples without replacements.\n\nso the probability that shipment is not rejected is:  Probablilty of not picking up defected product the first time (95 out of 100) time prob of not picking second time (94 out of 99) etc\n\n95/100 * 94/99 * 93/98 * 92/97 * 91/96 *90/95 * 89/94 * 88/93 * 87/92 * 86/91\n\n= 0.584\n\nThe probabiliyt of shipment being reject is just 1- P(not rejected)\n\n= 0.416
735064
That sounds familiar... Think I'm working on it now! But I don't remember q. It is n in my book. In which case, q (if I'm thinking right) is the number of samples. Then again what's up with the ^'s? If np was > 10, then it meant that the distribution was approx. normal. You might be ahead of me right now...
1119536
1) If no one is injured, then no subs ever get to play.  The probability is (.9)^11 = .3138106 or about 31.38%.\n\n2) (.9)^9 * (.1)^2 = .003874205 or about 0.39%\n\n3) "At least" is best calculated as the complement of "no substitutes were needed."  In the first calculation we determined the probability of not needing any substitutes.  So the probability of needing at least one sub would be .6861894 or about 68.62%\n\n4) The trick here is that the two injured could both be players, or one could be a player and the other could be the sub.  (They could not both be subs, however.)  The calculation in #2 (above) is part of this--the probability that two of the players were injured.  For the other part we need to know the probability that one player is injured (while 10 are not) and that the sub is also injured.  .1 * (.9)^10 * .15 = .0052302.  So the overall result for either of these outcomes (add this result with the result from #2) is 0.00910438 or about 0.91%.
249369
Is the term you are looking for "probability"?\n\nI hope this helps.
907438
I could tell you the answer or I could give you an idea of what to consider when you do it yourself.\n\na) Binomial numbers and p = 0.7, q= 0.3 might help here.  You do not know the size of the population of the mailing list so you might need to make an assumption that it is large enough to assume a Normal distribution.  I think the central limit theorem might help here.\n\nb) Binomial theory, and conditional probabilities.  Look it up in your text book!  Again if the size of the population is large enough you might be able to simplify the problem.
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!)
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.
809537
I hope it's ok this way:\n\nHi John,\n\nHow are you? Please find below my concerns regarding the shipment. I need some clarifications.\n\nFirstly, What shipment method do you prefer for the shipment of Part No A in Po 10?\n\nSecondly, I found that the goods are going to be shipped via various shipment methods. The total size might not be too big. Would you like us in this case to ship all the goods by Air-freight or UPS?\n\nI am sorry that Part No A will be shipped with a shortage of 100 pieces and part No B with 200 pieces. This is because there has been an unexpected yield loss during production. We’ll try to complete the rest as soon as possible.\n\nOur priority is to meet your satisfaction, so could you please reply me before shipment date on Nov. 16, 2006.\n\nBest Regards,\nPeter W
922757
a) Can't sell more than 4, so 1*p(selling 1) + 2*p(selling 2) + 3p(selling 3) + 4p(selling4).\n\nb) Find the number n so that p(x>n) is less than 0.05 for a Poisson "vif" mean 3.
242742
Remember, chance events are not dependent on the previous outcome.  The correct answer, as others have pointed out, is 50%.
1099243
No. This is what we call probability. Take a statistics course.
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.
633627
Sorry for repeating some of the previous answers.\n\n(1) The probability of getting no aces (~X) is:\n\nP(~X) = (48/52)*(47/51)*(46/50)*(45/49)*(44/48) = 35673/54145\n\nso:\n\nP(X) = 1-P(~X) = 18472/54145\n\n(2) I agree with mathgirl826. The last reply is almost correct except that there are 4*3 = 12 possibilities for the selection of the two suits (not 6). Once the suits are selected there are C(13,3)*C(13,2) hands, so:\n\nP(Y) = 12*C(13,3)*C(13,2)/C(52,5) = 429/4165\n\n(3) blahb31's analysis looks good, but maybe a slightly easier way to do it is to first calculate P(Y|~X), the probability of Y when no aces are drawn. It's just the same as P(Y) would be if the deck contained only 48 cards with 12 cards (no aces) in each suit:\n\nP(Y|(~X)) = 12*C(12,3)*C(12,2)/C(48,5) = 110/1081\n\nThen we can use (this follows from P(X and Y) = P(Y) - P((~X) and Y)):\n\nP(Y|X)\n= P(X and Y)/P(X)\n= (P(Y) - P(~X)*P(Y|(~X)))/P(X)\n\nAll of the values in the above expression have already been calculated:\n\n= (429/4165 - (35673/54145)*(110/1081))/(18472/54145)\n= 1947/18472\n\nThis is about 2% higher than P(Y). The event Y which indicates a relative lack of diversity in suits, is somewhat correlated with a decreased probability of duplication in rank values, and therefore a slightly higher probability of finding any given rank such as an ace.
656922
Okay, intially, you pick one marble, out of 12. and there are 3 yellows, then your initial chances are 1/4 of picking a yellow. Now, you've picked one, so there are 11 marbles left in the bag. There are 4 blue marbles in the bag, so you have a 4/11 chance of picking a blue ball at tthis point. \nTo Calculate Probability based  on an event that 1) Has already happened ( the yellow ball was picked first) and that is an event depending on the results of the , you simply multiply the chances. ( 1/4*4/11) or 4/44, or 1/11 chances that that will happen.
726292
I believe you just gave the probability of the favorable outcome: 3/5.  You need to phrase your question in clearer terms.  More Web 1.0, less Web 2.0.
600596
There are 6 possible outcomes for the roll. There is only one outcome when the number is 4. Therfore the chances of getting a 4 are 1/6.
877981
For the first part, per hour the rate is 30.  so we have lambda = 30, so we know the mean and variance (look at the answer before mine).  To find the Probability the 240th person enters in the 8th hour, find the probability 240 or more people arrive in an 8 hour day, then times that by the probability less than 240 arrive in 7 hours.\n\nso (1-P(<240 in 8))*P(<240 in 7)\n\nRemember, The poisson process is memoryless.\n\nSteve