this is for people how like probability questions (i need help fast)?
354101
This is easy, just say there are 100 shaft holders.  So of A, 50 of them are processed there, and 3% don't work, so 1.5 out of 50.  For B, it is 30 x 4%=1.2, and C is 20 x 5%=1.  \n\nSo out of a total 100, 3.7 don't work, the the probability that this one chosen at random wont work is 3.7%.  \n\nHopefully the way i calculated the answer helps you out in the future.
277670
probability is the # of favorable outcomes over the # of possible outcomes:\n\nprobability=\n# of favorable outcomes\n------------------------------\n# of possible outcomes\n\nfavorable outcomes would be the # of possible resultis in your favor\n\nexample:\ni have 5 cards- a joker, a jack, a queen, a king, and a ace. there is a 1 in 5 chance (1/5) that you will pull a jack...there is a 3 in 5 chance (3/5) that you will pull a royal card (jack, queen, king)\n\nin both examples, the first number (1 and 3) are favorable outcomes
475811
Here's a step by step explanation of how the probability is assessed.  Hope this will make things clearer for you.\n\nhttp://www.mste.uiuc.edu/java/java/birthday/explanation2.html
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.
855670
Boy, this is tricky.  I think the probability is (8! * 41c2 * 10c2)/49!, where 41c2 = 41!/39!2! and 10c2 = 10!/8!2!  In this number, 8! represents the correct picks, 41c2 the incorrect picks, and 10c2 the number of ways to order the picks.\n\nFor exactly seven correct numbers in 10 picks , only a minor change is needed.  It becomes (7! * 42c3 * 10c3)/49!.  Follow this pattern for 6 of 10, 5 of 10, etc
245215
There are five tosses with two choices each.\nEach toss is a 50% chance of coming up heads regardless of the results before it or after.\nThe probability of getting a certain number of heads would be calculated as such: number of heads/number of choices.\nWe are looking for two heads and there are ten choices (heads and tails of each coin)\nThe probability of it would be 2/10 or 20%\n\nIt is correct in that each toss is 50% chance, but in this case we are speaking of of the chances of a group of coins having a certain number turn up heads.  We are also speaking of the probability of this before any actions are taken.  The odds will change with each coin toss till the last one which will be 50%
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.
631161
This is a binomial probability.\nP(success) = 1/5, P(failure) = 4/5.  There are 6 events here, so P(2 successs) = (1/5)(1/5)(4/5)(4/5)(4/5)(4/5).\nP(at least 2 successes) = P(2 successes) + P93 successes) + . . . . P(6 successes) or 1 - P(O successes) + P(1 success) + P(2 successes) which you already computed above.\nLet me know if you don't understand.
147164
Ah, a variation on the Monty Hall problem:\n\nhttp://en.wikipedia.org/wiki/Monty_hall_problem\n\nIt is a probabilistic curiosity, this one. The highly unintuitive answer is that you actually **can** improve your chances by changing your mind.\n\nYour problem appears to only be different in that it starts with 5 boxes. The above link should give you sufficient information to solve the problem.
955894
x1*p(x1) + x2*p(x2)\n\nx1 = 1 for  event dots> 3,\nx2 = 4 for event  dots<=3.  \nThese are equally probable because there are 6 dots on the die.  \n\n1*(1/2) + 4*(1/2)
131088
try this site\n\nhttp://www.futureaccountant.com/probability/
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.
560810
1)  I think you're missing a bit.\n\nGiven that P(W<=4)=0.825 then P(W>4) = 1-0.825 = 0.175\n\nThis is because all probabilities sum to 1.  So since <=4 annd >4 make up all teh possibilities the value could take - then they're total probabilties must be 1.\n\n2)  Expected value\nTo do these you simply have to multiply the probability by the value and total them up.\n\n+36,000×0.7 -6,000×0.3 = $23,400\n\n3)  Binomial\nYou have a bin(6, 1/6) as there are 6 trials and the probability of a success (ie getting a four) is 1/6.\n\nYou require the probability of 2 fours so just using the binomial formula P(X = x) = nCx × p^x × (1-p)^(n-x) you get:\n\nP(X = 2) = 6C2 × (1/6)² × (5/6)^4 = 0.2009\n\n4)  Binomial\n\nthis again is a binomial Bin(10, 0.33) - as we have 10 trials (students) and the probability of each one being a success (ie member of ethnic minority) is 0.33\n\nUse the formula above to get the probability of more than 6:\n\nP(X = 7) = 10C7 × 0.33^7 × 0.67^3 = 0.01538...\nP(X = 8) = \nP(X = 9) = \nP(X = 10) = \n\ntotalling these up gives the probability of more than 6.\nThen the probability of NOT more than 6, P(X <=6) = 1 - P(X>6) using the idea from 1.\n\n5)  The mean of teh binomial is just np (ie the number of trails time the probabilty of success).\n\nSo for your example where we have bin(6, 0.7) it will be 6×0.7 = 4.2 seeds germinating.
917829
When you are computing the probability of a given event such as 2 heads, 2 tails, or one of each, you are essentially counting the number of ways a given event can occur and dividing this number by the number of all possible outcomes.\n\nIf you toss 2 coins once ( or one coin twice) there are 4 possible outcomes:\n\nHH = both heads\nHT = 1st heads, 2nd tails\nTH = 1st tails, 2nd heads\nTT = both tails\n\nAssuming the coins are "fair" each outcome is equally likely.\n\nThere is one way in four that two heads (or 2 tails) can happen so the probability for each of these outcomes is 1/4\nBut there are two ways you can have one head and one tail and so the probability of this happening is 2/4 =1/2. The sum of all the probabilities of the distinct outcomes will always equal 1. \n1/4 +1/4 + 1/2 =1\n\nHow about if you have 3 tosses? The 8 possible outcomes are as follows:\n\nHHH \nHHT\nHTH\nTHH\nHTT\nTHT\nTTH\nTTT\nWe see there is only one way to have 3 heads (or 3 tails) and so the probability of 3 heads (or 3 tails) = 1/8\nHowever there are 3 ways you can have 2 heads and one tail, so the probability of this happening is 3/8. And there are also 3 ways you can have one head and 2 tails, so 3/8 is the probability for this too.
919178
.................... / ....... &#92;\n................ 0.85 ..... 0.15 (Pass test 1)\n.............. / ....... &#92;\n.......... 0.85 ..... 0.15 (Pass test 2)\n........ / ....... &#92;\n... 0.85 ..... 0.15 (Pass test 3)\n(Fail all 3)\n\n1) How many outcomes does your tree show?\n\nThere are four outcomes:\nP(Pass Test 1) = 0.15 = 15%\nP(Pass Test 2) = 0.85 * 0.15 = 12.75%\nP(Pass Test 3) = 0.85 * 0.85 * 0.15 = 10.8375%\nP(Fail all 3) = 0.85 * 0.85 * 0.85 = 61.4125%\n\n2) What is the probability she will eventually pass an exam.  \n\nThis is the sum of all outcomes where she passes:\nP(Pass Test 1) + P(Pass Test 2) + P(Pass Test 3)\n= 38.5875%\n\nAlternatively you could take the 1 - P(Fails all 3) and get the same answer.\n\n3) What is the probability she will take the exam three times?\n\nIt didn't say whether she passes or fails, so the chance she has to take the test a third time is 0.85 * 0.85 = 72.25%\n\n4) What is the expected number of times she will take the exam?\n15% of the time, she'll take it once\n12.75% of the time, she'll take it twice\n72.25% of the time, she'll take it three times.\n\nSo the expected number of times is:\n0.15 * 1 + 0.1275 * 2 + 0.7225 * 3 = 2.5725\n\nSo she's expected to take it 2.5725 times.\n\n5) What is the expected number of times she will fail the exam?\n15% of the time she will pass test 1 (failures = 0)\n12.75% of the time she will pass test 2 (failures = 1)\n10.8375% of the time she will pass test 3 (failures = 2)\n61.4125% of the time she will fail all 3 (failures = 3)\n\n0.15 x 0 + 0.1275 x 1 + 0.108375 x 2 + 0.614125 x 3 = 2.186625\n\nSo she's expected to fail it 2.186625 times.
540053
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
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!)