Does anyone know how to find the expected value for this probability distribution?
337735
In this case, a 70% success rate is independent of surgical history. So for nine patients, you can expect 6.67 (rounded to  7) of them to have successful surgeries. The rate of success is not cumulative at all.
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
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
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.
781591
P(wrong size) = 0.2\nP(defective) = 0.35\nP(not defective and right size) = 0.60\n\nThus P(defective or wrong size) = 1 - P(not defective and right size) (These two events are opposites.  Look at the Venn diagram for help with this, it's DeMorgan's law if you're curious)\n= 1 - 0.60 = 0.40\n\nP(defective or wrong size) = P(defective) + P(wrong size) - P(defective and wrong size)\n\nSo 0.40 = 0.2 + 0.35 - P(defective and wrong size)\n0.40 = 0.55 - P(defective and wrong size)\nP(defective and wrong size) = 0.55 - 0.40 = 0.15
341707
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
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.
19818
try\nhttp://www.peterwebb.co.uk/probability.htm
584200
1) the pdf is actually a binomial.\nIf you consider only 1 man, what is the probability that he will be paired to his wife out of the 10 women? \nIt is 1/10\n\nLet X be the number of paired couples, paired meaning they are paired to their own spouse.\nTherefore X~B(10,1/10)\n\nSo if we want to find the probability that there is only 1 pair, using the formula of a binomial dist,\n\nP(X=1)= [10 choose 1] (1/10) (9/10)^9\n= 0.3874\n--------------------\n\n2) To find the mean and variance, using the Binomial parameters formula, we compare what we got in part 1 with the general formula,\nX~B(n,p)\nX~B(10,1/10)\n\nSo our mean = (n) multiply (p)\n= 10 x (1/10)\n= 1\n\nVariance = (n) multiply (p) multiply (1-p)\n= 10 x (1/10) x (9/10)\n9/10\n----------------------\n3) P(X >= 3) = 1 - P(X <= 3) \n= 1 - [ P(X=0) + P(X=1) + P(X=2) + P(X=3) ]\nreferring to the above eg in part 1, its easy to derive\n= 1 - [ 0.0349 + 0.3874 + 0.1937 + 0.0574]\n= 0.3266
795924
Binomial distribution, (a) p=0.5, n=12; (b) p=0.5, n=13.  You can do it!
131088
try this site\n\nhttp://www.futureaccountant.com/probability/
928594
If you assume it's normal, then yeah, although I assume there is more data than you have given as you do not have the variance there, although you could treat this as a binomial distribution giving it mean 0.3, variance n0.3*0.7, although when calculating the confience interval you'll want to divide by n^2 for that variance.
947255
Let X be mean fill of container. \n\nThe distribution is X~N(16, 1.5) i.e. it is normally distributed as stated in the question.\n\nWant to find P(X>17), so standardise.\nP(X>17) = P(Z>(17-16)/(1.5/2)) = P(Z>1.33)\n\nThen you look up your normal table. The answer is approximately 0.09.
799460
Figure the probability of each outcome first:\nP(red marble) = 4/18 = 2/9\nP(green marble) = 3/18 = 1/6\nP(white marble) = 5/18\nP(black marble) = 6/18 = 1/3\n\nThe expected winnings would just be:\n2/9 * $5 + 1/6 * $4 + 5/18 * (-$3) + 6/18 * (-$2)\n\nThis works out to an expected winning of $0.277778.\n\nAn alternate way to figure this is:\n[(4 * $5) + (3 * $4) + (5 * -$3) + (6 * -$2) ] / 18\n= (32 - 27) / 18\n= $5/18 = $0.277778\n\nThe game is better than fair, since you don't have to pay anything but you are expected to win more than $0.  (Technically it is unfair since the expected return is not equivalent to what you pay, but I'm going to keep playing it because it is unfair to the house, not to me!)
508493
I've found that the Wikipedia articles on probability distributions are top notch and contain enough information for you to solve this problem on your own.\n\nHint: solve the algebraic expressions that relate mean and variance so you can express p or n as algebraic relationship of one another and then do some substitions.\n\nAt first I thought you needed to know p or n up-front however I realize that you have all the info you need to solve this problem algebraically.\n\nGood luck with your homework.
417168
Let E be the event that a man picked up randomly (with uniform distribution) has that kind of wallet.\n\nP(E)=1000/2000000=1/2000 \n\nLet G be the event that the man is the thief.\n\nP(E|G)=1 (Assuming that the thief certainly has that wallet)\n\n(i) P(G|E)= P(G,E)/P(E)=P(E|G)*P(G)/P(E)\n=P(G)/P(E)=2000*P(G)\n\n(ii) P(G)=1/1999999 => P(G|E)=2000*1/1999999 &#92;approx 1/1000\n\n(iii) Either a decreased P(E) or an increased P(G) will make P(G|E) higher. P(E|G)=1 is not a realistic assumption since it is likely that the thief will get rid of the wallet with some probability.
418035
Abby's answer is for getting exactly \nTTTTTTTTHH\n\nBut the same probability applies for\nHHTTTTTTTT\n\nand\n\nHTHTTTTTTT\n\nThere are C(10,2) ways the 8 tails and 2 heads can occur, so the correct answer is to multiply .5^10 * C(10,2)\n\nThis is the same as c(10,8)/ 2^10 as in the last answer.