A) Suppose that a radio tube inserted into a certain type of set has probability of 0.2 of functioning more than 500 hours. If we test 20 tubes, what is the probability that exactly 5 of these functions more than 500 hours.\n\n\nB) A continuous random variable X has the following probability density function:\n\nf(x) = Ax^2 (1 - x), 0 < x < 1\n= 0, otherwise\n\nDetermine the constant A and find\np (1/3 < X < 1/2) .
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For A), use the Poisson distribution. M. Poisson was a French mathematician whose name also happens to mean "fish" in French. To those who disagreed with him, M. Poisson said, "One man's meat is another man's Poisson!" \n\nIf you tested 20 tubes many times, on average 4 of them would last more than 500 hours (.20 of 20). The formula you need for the probability of exactly 5 lasting more than 500 hours is\n\nP(5) = e^-4 * (4^5/5!)\n\nThis is not so hard to compute. 4^5 is 1024, and 5! is 120. So we have\n\nP(5) = e^-4 * 1024/120\n\nP(5) = 0.018 * 1024/120 = .156\n\nFor B), I don't whether (1 - x) is in the exponent or is a factor of multiplication. Please clarify with parentheses or asterisks.\n\nf(x) = Ax^[2 * (1 - x)] ?\n\nor\n\nf(x) = A * (1 - x) * x^2 ?\n\nI see that math girl is using the latter, wherein (1 - x) is a factor of the multiplication. In that case, I agree with her that A = 12, for the reason she gives that the integral from 0 to 1 is equal to 1. She gives you the expression for the integral so you should be able to find the value of it over the interval 1/3 to 1/2. If not, choose this answer as best and ask me to integrate your expression from 1/3 to 1/2 for you.
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If a random sample size of 25 corresponds to a standard deviation of 30, then you can expect a random sample of size 25N to roughly correspond to a standard deviation of 30 / sqrt(N).\n\nIf 99% of the area under the bell curve is between -Z and +Z standard deviations, then you want 30 Z / sqrt(N) to be less than or equal to 0.5.\n\nMaybe a little more because of the variation of the estimated standard deviation from the true standard deviation.\n\nAlso check that the theory and the measurements both suggest it is appropriate to use the normal or Gaussian distribution.
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This is a good question, and is the basis behind the theory of statistical thermodynamics. \n\nWhen you roll one die (i'm assuming a standard, fair,  six sided die) the probability of getting any particular number is 1/6. That is pretty obvious, each number is equally likely.\n\nInteresting things start to happen when you roll multiple dice however. Lets take all the possible outcomes for rolling two standard dice.\n\nDie 1   -   Die2   -   Total\n  1          1           2\n  1          2           3\n  1          3           4\n  1          4           5\n  1          5           6\n  1          6           7\n  2          1           3\n  2          2           4\n  2          3           5\n  2          4           6\n  2          5           7\n  2          6           8\n  3          1           4\n  3          2           5\n  3          3           6\n  3          4           7\n  3          5           8\n  3          6           9\n  4          1           5\n  4          2           6\n  4          3           7\n  4          4           8\n  4          5           9\n  4          6           10\n  5          1           6\n  5          2           7\n  5          3           8\n  5          4           9\n  5          5           10\n  5          6           11\n  6          1           7\n  6          2           8\n  6          3           9\n  6          4           10\n  6          5           11\n  6          6           12\n\nNow each die is independent of the other one. This means that getting 1-6 is a different result to getting 6-1. Thus it's easy to see that there are 36 possibile outcomes as shown above (6*6). Take a look at the totals of the outcomes though.\n\nThe first total i've written down is '2'. This comes from getting a 1 and another 1. You'll notice that this total does not appear anywhere else at all. The only possible way to get a total of 2 is if you get a 1 on both dice.\n\nAlso look at the last total, '12'. This is from getting a 6 and another 6, and is also the only place you'll see this total. There is only 1 possible way to get 12 as a total.\n\nHowever if you look at the total '7', you'll notice this appears 6 times. There are 6 different ways to get a 7 (1-6, 2-5, 3-4, 4-3, 5-2, 6-1). Therefore you are 6 times more likely to get a total of '7' than you are to get a total of either '1' or '12'! If you were to plot the average score received after rolling the two dice and plot them on a probability distribution curve (just graph the average scores along the bottom with the probability of getting that score up the side) you will see that it starts of with a low probability for getting an average of 1, rises to a peak at getting an average of 3.5, and then drops back to a low probability got getting an average of 6.\n\nNow with two dice there is still a reasonably good chance of scoring an average of 1, and a reasonably good chance of scoring an average of 6 (you'll see it happen every now and then if you roll the dice for a few minutes), but what happens if we go to 3 dice? Well the number of possibile outcomes is multipled by 6 (don't worry, I won't list them all!) But you'll notice there is STILL only 1 possible way of getting an average score of '1' (1-1-1). And there is STILL only one possible way of getting an average score of '6' (6-6-6). Therfore, the probability of you getting these numbers as an average of rolling 3 dice is much less than with two. Also the probability of getting an average of 3.5 is much higher (i'll leave that to the reader to confirm). If you plot the results for 3 dice onto the probability curve you'll notice that it is a bit steeper. It still has it's peak at 3.5, but it is much sharper.\n\nThe really interesting things happen when you go to large numbers of dice. Imagine that you have one die for every atom in a sample of some substance. Ther
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when the probability is 0.2:\n\nP((β-12%)*Sqrt(n)/σ>=1.301)=20%\n1.301 is the T value under degree of freedom of 44 and α =10%\n\nI did not pay attention in leaasons of statistics, I dont know if it's correct.
7321
try\nhttp://www.peterwebb.co.uk/probability.htm
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Part (a) of problem 1 is pretty easy.  There are 10 marbles total, and 6 of them are red.  On your first draw, you have a 6/10, or 0.6 (60%) probability of drawing a red sock.  In other words, there are 6 ways of drawing a red sock out of a total of 10 possibilities.\n\nFor part (b), you need to know that the probability of two independent events happening is given by the product of the probabilities of each of the individual events.  In this case, the two independent events are that the first marble drawn is red AND the that the second marble is also red.  We already calculated the probability for drawing a red marble on the first draw in part (a).  After the first draw, the box contains 9 total marbles, 5 of which are red, so the probability that the second marble is red is 5/9.  The probability that both are red is given by the product, 6/10 * 5/9 = 30/90 = 0.3 (30%).\n\nThe reasoning for part (b) is the same as for part (a).  The probability that the first marble drawn is white is 4/10, and the probability that the second marble drawn is also white is 3/9.  The probability that both are white is given by the product, 4/10 * 3/9 = 12/90 = 0.133... (13.333...%).\n\nFor part c, there are two ways one can get a red and a white marble: either the first marble is red and the second white, OR the first marble is white and the second red.  When there is more than one way to achieve the desired result, the probability of getting that result is given by the SUM of the probabilities of each of the different ways to achieve that result.  In this case, the probability of drawing red-white is 6/10 * 4/9 = 24/90, while the probability of drawing white-red is 4/10 * 6/90 = 24/90.  The total probability of getting a red and a white marble is then 24/90 + 24/90 = 48/90 = 0.53333....\n\nUsing the same reasoning, you should now be able to solve parts (a) and (b) of the second problem.  You should find for part (a) that the probability is equal to 0.42424242...., and that the probability for part (b) is 0.09090909....\n\nFor part (c), you would use a similar logic as used for part (c) of the first problem.  The probability of getting a matched pair (either both white or both blue) is given by the sum of the probabilities found in parts (a) and (b).  You should find that the probability is 17/33 = 0.51515151....\n\nFor the third problem, you need to know that the expected "payoff" of a game is given by the sum of the products of the prize value times the probability of winning that prize.  In this case, the probabilities of winning the various prizes are:\n\n$100 : 1/500\n$50 : 3/500\n$25 : 5/500\n\nSo the expected payoff (i.e., the "mean" or "average" winnings for a single ticket) is given by:\n\nP = $100*(1/500) + $50*(3/500) + $25*(5/500)\nP = $100/500 + $150/500 + $125/500\nP = $375/500\nP = $0.75
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(a) True  because 460-393 > 0\n(b) False, heat of formation is 0\n(c) False, =3X393\n(d) False, 436.4 > 393\n(e) False, = 460
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the Chebyshev's inequality goes Pr(|X-mean| > or = k.sd)< or = 1/k^2\n\ni m assuming u meant standard deviation when u said 4.3818, which is 2 sd.\n\nThe prob of points outside 2sd at both sides is not more than 1/k^2.. or 1/2^2=1/4.\n\nSo, the probability of points within 2sd will be at least 1-1/4=3/4.\n\nhow many data points? if u have 100 data points in total, there will be at least 100x3/4=75 data points within the mean+/-4.3818 range.\n\nNote: The exactly probability depends on the type of distribution the data follows, for example, for Normal/Gaussian, 95% lies within +/- 1.96sd. Chebyshev's inequality states the minimum possible for any type of distribution.
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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
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The answer is 1/3.\n\nAnswer 1 way: Given that it wasn't a club, then it must have been 1 of the 3 remaining suits all with equal likelihood.\n\nAnswer 2nd way: P(A|C) = P(A)/P(C)\n\nIn this instance C = ~B (not B) = 3/4\n\nStandard P(A) = 1/4.  \n\nSo, the conditional prob is (1/4)/(3/4) = 1/3.
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The RAND() function is not the appropriate way to solve this.\n\nThe number of cars can be determined a couple of ways.\n\n1.  One is to use NORMINV(.99,2600000,100955) = 2834856\n2.  Or you could use the standard normal distribution:\n\n    x= 2600000 + 100955 NORMSINV(.99)\n     = 2600000 + 100955(2.326348) = 2834856\n\nRAND() just gives you a random number between 0 and 1.  I don't see how that would help you in this problem.\n\nThe second problem is also easily solved using the the standard normal distribution:\n\nFirst, the 2.4 million is 200000 below the mean, \nso z=-200000/100955= -1.98108\n\n2.7 million is 100000 above the mean, so \nz=100000/100955=.99054\n\nSo the probability the number of cars will be between 2.4 and 2.7 million is\n\n= NORMSDIST(.99054) - NORMSDIST(.023791)\n=.839045 - .023791 = .815254
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The sample space consists of all pairs (x,y) with 1 <= x <= 6 and 1 <= y <= 7. There are 6 * 7 = 42 in elements and each is equally probable. You only need to count up the number of elements in each event (subset of the sample space) and divide by 42 to get the answers:\n\n(a) 1 (1,1) --> P = 1/42\n(b) 6 (2,x)'s +  5 (x,2)'s --> P = 11/42\n(c) 9 (even,even)'s --> P = 9/42\n(d) 12 (even,odd)'s + 9 (odd,even)'s --> P = 21/42 = 1/2
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Whats the probability that the problem is *not* solved? Tom doesn't solve it (0.75) and Alice doesn't (0.5) so its 0.75*0.5 = 0.375. So the probability it is solved is 1-0.375 = 0.625.\n\nNow, we are given that it is solved. What is the probability that Tom solves it? 0.25. \nSo:\nP(Tom | solved) = P(Tom and solved) / P(solved) = P(Tom) / P(solved) = 0.25/0.625 = 0.4.\n\n(edit - note that the reason the below answer is not quite correct is because it is possible for *both* of them to solve it - so the probability Tom solves it when solved + probability Alice solves it when solved should *not* equal 100% - it should be more.)
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49/99\n\nProof:\n\n1) Probability that first passenger (P1) sits in last passengers (P100) seat (S100) is 1/99\n2) probability that P1 sits in S99 , and P99 sits in S100 is (1/99) * (1/2)\n3.1) probability that P1 sits in S98, and P98 sits in S100 is (1/99) * (1/3)\n3.2) probability that P1 -> S98, P98->S99, P99->S100 is (1/99) * (1/3) * (1/2)\n4.1) P1->S97, P97->S100 = (1/99)*(1/4)\n4.2) P1->S97, P97->S99, P99->S100 = (1/99)*(1/4)*(1/2)\n4.3) P1->S97, P97->S98, P98->S100 =  (1/99)*(1/4)*(1/3)\n4.4) P1->S97, P97->S98, P98->S99, P99->S100 = (1/99)*(1/4)*(1/3)*(1/2)\n5.1) P1->S96, P96->S100 = (1/99)*(1/5)\n5.2) P1->S96, P96->S99, P99->S100 = (1/99)*(1/5)*(1/2)\n5.3) P1->S96, P96->S98, P98->S100 =  (1/99)*(1/5)*(1/3)\n5.4) P1->S96, P96->S98, P98->S99, P99->S100 = (1/99)*(1/5)*(1/3)*(1/2)\n5.5) P1->S96, P96->S97, P97->S100 = (1/99)*(1/5)*(1/4)\n5.6) P1->S96, P96->S97, P97->S99, P99->S100 = (1/99)*(1/5)*(1/4)*(1/2)\n5.7) P1->S96, P96->S97, P97->S98, P98->S100 =  (1/99)*(1/5)*(1/4)*(1/3)\n5.8) P1->S96, P96->S97, P97->S98, P98->S99, P99->S100 = (1/99)*(1/5)*(1/4)*(1/3)*(1/2)\n\n1) (1/99)\n2) (1/99)(1/2)\n3) (1/99)(1/3)(1 + ½)\n4) (1/99)(1/4)(1 + 1/3)(1 + ½)\n5) (1/99)(1/5)(1 + ¼)(1 + 1/3)(1 + ½)\n\ni) (1/99)(1/i)product(j=2 to (i-1),(1+1/j))\n\n99) (1/99)(1/99)product(j=2 to 98,(1+1/j))\n\nGrand Sum = (1/99)*(1 + ½ +sum(i=3 to 99,(1/i)*product(j=2 to (i-1),(1+1/j))))\n=(1/99)*(1 + ½ + sum(i=3 to 99,(1/2)))\n=(1/99)*(1 + 98/2)\n= 50/99\n\nProbability that last passenger gets bumped = 50/99\n\nHence Probability that last passenger does NOT get bumped = 1 – 50/99 = 49/99\nA bit less than 50%\n\nWould be 1/2 like others suggest if first passenger was not constrained to sit in the wrong seat as given in the problem statement.
432902
a) Both: .1(.02) = .002\n\nb) Exactly one:  .02(1-.1) = .018 + .1(1-.02) = .018 + .098 = .116\n\nc) (1-.02)(1-.1) = (.98)(.9) = .882\n\nThey add up to one.  I'm not a probability person, but I think this is correct.
73866
Probability that the first coil is good = 88/100\nProbability that the second coil is good = 88/100 (irrespective of the first choice)\n\nProbability that both are good = (88/100)^2 = 77.44%
373483
The second problem is easier than the first problem, so I'll tackle it first.\n\nLet's imagine that the trip is 120 miles. The distance doesn't matter if the average rates are as given above.\n\nIt takes 3 hours to travel 120 miles at 40 miles per hour.\nIt takes 2 hours to travel 120 miles at 60 miles per hour.\n\nSo it took 5 hours to travel a total of 240 miles.\n\nThis is an average rate of 48 miles per hour.\n\nI'll work on the first problem and hopefully I'll get back to you.\n\nUPDATE\n\nI think I have the first problem solved as well. It involves an infinite series! Call the people involved A and B.\n\nThe probability that A wins on the first toss is 1/2.\n\nIf A doesn't win on the first toss, then the probability that B wins on the second toss is also 1/2. But this only happens in 1/2 of all games. (In the other half, A won on the first toss.)\n\nIf B doesn't win on the second toss, the the probability that A wins on the third toss is also 1/2. But this only happens in 1/4 of all games (In half the other games, A won on the first toss. In half the remaining games - 1/4 of all games, B won on the second toss.\n\nFollowing this logic, A wins:\n\n1/2 + 1/4(1/2) + 1/16(1/2) + 1/64(1/2) + ...\n\nof all games.\n\nB wins:\n\n1/2(1/2) + 1/8(1/2) + 1/32(1/2) + ...\n\nof all games.\n\nLooking again at A, we have:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...)\n\nThe term in parentheses is an infinite series having initial term 1 and ratio 1/4.\n\nThe sum of an infinite series having ratio whose absolute value is less than 1 is equal to:\n\na/(1 - r)\n\nWith a = 1, and r = 1/4, the sum = 4/3. So:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...) = 1/2 * 4/3 = 2/3\n\nLooking again at B, we have:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...)\n\nThis sum should be half of the sum of the infinite series in A.\n\nWith a = 1/2 and r = 1/4, a/(1 -r) = 2/3\n\nSo:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...) = 1/2 * 2/3 = 1/3.\n\nA should win 2/3 of the time, and B should win 1/3 of the time. There is an advantage to going first!