hard probability math ?
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.
1076427
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.
407360
yea, multiplying the single event probabilities of two independent events to get the prob of the independent events conjunction is correct. In other words (to avoid you going to grab an Advil already): p(A&B)=p(A)*P(B) assuming A and B two independent events. That's BASIC stats.\n\nIn excel, use the function POWER. First argument the probability, second argument the number of occurence in a row.
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%.
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/
292028
For and event, odds are ususally expressed as the number of possible outcomes that qualify as a success compared to the total number of possible outcomes: 1 in 6, 3 in 10, etc.\n\nProbability is the same thing expressed as a fraction:\n1/6 = .66... 3/10 = .3\n\nWhen you select k objects from a set of n objects, the number of possible ways that you can do this is n!/(k!(n-k)!).\n\n\nIn all of your problems, your event is selecting a set of 3 objects from a set of 16. The number of possible outcomes of this is: \n16!/(3!13!) = 16x15x14/6 = 8x5x14 = 560.\n\nThe number of outcomes that are considered to be a success are:\n\n3 forks (pick 3 forks from the set of 7) = 7!/(3!4!) = 7x6x5/6 = 35\n\n2 forks 1 knife  (pick 2 forks from the set of sevan and 1 knife from the set o 5) = 7!/(2!5!) x 5 = 7x6/2 X 5 = 21x5 = 105\n\n3 spoons (Select 3 spoons from the set of 4) = 4!/(3!1!) = 4\n\n1 fork, 1 knife, 1 spoon  (Select 1 fork from the set of seven, 1 knife from the set of 5 and one spoon from the set of 4) = 7!/(1!6!) x 5!/(1!4!) x 4!/(1!3!) = 7 x 5 x 4 = 140\n\n1. P(3 forks) = 35/560\n2. P(2 forks, 1 Knife) = 105/560\n3. P (3 spoons) = 4/560\n4. P(1 fork, 1 knife, 1 spoon) = 140/560
624712
This problem is solvable.\nProbability of event A is written P(A).\n\nP(A)= Total # of outcomes that correspond to event A (divided by -or- over) Total # of outcomes in sample\n\nAnswer:\nP(A) = 12/20 = 3/5
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
664305
For that to happen, 1 must come on each dice.\n\nThus P(Sum is 2) = P(1 appears)*P(1 appears)\n\n=(1/6)^2 = 1/36
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.
279356
This is always so much fun, isn't it.  This is probably why the previous answerer is now looking for satisfying employment in the food-service sector.\n\nBy the way, you forgot to mention if they have to specifically be in that order or in any order, though.  It's a very different answer.\n\nOkay, what's the probability of drawing one green marble each time you draw, if you put the marbles back each time?  It's the partial over the sum, or in other words 4 out of 6, or 2/3.  Similarly the probability of drawing a red marble is 2 of 6, or 1/3.\n\nSo if you specifically want the order to be green, green, red, then you just multiply the probabilites each time.  2/3 * 2/3 * 1/3 is  4/27.\n\nHowever, suppose it can be in any order?  Here's where it gets tricky.  First we figure how many ways there are to draw two greens and a red -- we can draw (g,g,r) or (g,r,g) or (r,g,g) -- so three ways.  The probability is the same for each time so we add them together, essentially 3 * 4/27 is 12/27, or a little less than 50%.\n\nPart b) is harder, since the probabilities change with each draw.  The first time we draw it's a 2/3 chance of getting green, but after that there are only 3 green marbles out of 5, so it's now 3/5 instead of 4/6.  So if we want to draw two green and one red IN THAT ORDER it's 2/3 * 3/5 * 1/2 = 1/5.\n\nIf you don't care the order, then figure out each of the ways to draw 2 greens and one red (same as above) and add the probabilities from each time.  I'll leave it to you for the answer.\n\nActually there are very few interesting people in the world who like statistics.  But even the boring ones can usually find very good jobs.  Also it's been a while since my stats classes so if I'm wrong, please don't come after me with a baseball bat, thanks  :)
898818
There are sixty-six possible outcomes in the first two scenarios, and fifty-five possible outcomes in the third (Mary does not win):\n\nWhat is the probability a married couple wins the prizes? \n9.09% (6/66 x 100)\n\nWhat is the probability one prize goes to a man and one to a woman? \n54.5% (36/66 x 100)\n\nIf Mary does not win a prize, then what is the probability that John will? \n18.2% (10/55 x 100)\n\n(or 0% if John never made it to the party!)
254656
I'm assuming you are talking about a probability problem in Bayes' theory since you haven't specified what any of your symbols mean. This is why there are no responses.\nThe answer is actually quite clear. Draw three circles in a triple intersection for the events R, S and I and label it clearly. The diagram simplifies to a certain double intersection that motivates the result.
791422
Each individual toss is 1 / 2, a 50% probability.\n\nThe possibilities of all outcomes is 1 / (2 x 2 x 2), a 12.5% probability.\n\nOnly 2 possible outcomes do not meet the desired outcome (all heads, all tails).\n\n2 x 12.5% is 25% (but that's the opposite of what you want)\n\n100% - 25% = 75%
764469
X follows a Binomial distribution with n = 10 and probability of success p = 1/6
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.
152796
try this site...\n\nhttp://www.futureaccountant.com/probability/