1.  Let A and  B be events with P(A) = 3/8,  P(B) = 5/8 and P(A U B ) =  3/4. Find  P(A | B)  and P(B | A).\n\n2. Find P(B | A) if \n(i) A is a subset of B  and (ii) A and B are mutually exclusive events.
133513
I'm using && for intersection and || for union.\n\n1. P(A && B) = P(A) + P(B) - P (A || B) = 1/4\nP(A|B) = P(A&&B)/P(B) = (1/4)/(5/8) = 2/5\nP(B|A) = P(B&&A)/P(A) = (1/4)/(3/8) = 2/3\n\n2. (i) 1 because if A is a subset, then whenever A occurs, B must occur\n(ii) 0 because if A happens then B can never happen
235659
Start by multiplying both sides by D this gives you\n(D*P) = (2*s*t)\n\nthe divide both sides by 2t which gives you\n(D*P)/(2*t) = s\n\nor\n\ns = (D*P)/(2*t)
326188
The probability of A is 26%.  Probability of B is 42%.  Both A and B is 7%.  Probability of neither = 100% - [(26%+42%)-7%] = 39%.  Or in your format, 0.39.
507811
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.
215829
a) P(D+|K+) = P(D+ and K+)/P(K+)\n17/30 = P(D+ and K+)/.6\nP(D+ and K+) = .34\n\nb) P(D- | K+) = P(D- and K+)/P(K+)\n=P(D- and K+)/.6\n=(P(K+) - P(D+ and K+))/.6\n=.26/.6\n= 13/30
470300
well in my skool we take it as {<and rule>} and {<or rule>example:there is 5 white marble and 5 black marbles in a bag...u pick 2 randomly without replacing them in the bag.....wat is the probability that u get 2 white marbles==>this is a combination 2 the way it is solved is multiplying the the 2 combined probabilities\nanswer:pr:(white AND white)=(5/10)*(4/9)\n\nnow the set is this way....wat is the probability that u get 2 marbles of different colors in the first 2 picks===>set rule\n\nanswer:there is 2 combinations in which u can get 2 different colors there are {white then black} OR {black then white} u find the 2 combinations then ADD them\npr(WandB OR BandW)={(5/10)*(5/9)}+{(5/10)*(5/9)}
449608
1. An arithmetic sequence has a(sub 5)= -10 and a(sub 12) =18. Find a(sub 2) and a(sub 17)\n\nLet a(sub 1) = a and d = common difference.\nFor an AP:\na(sub n) = a + (n - 1)d\n\nSo a(sub 5) = a + 4d = -10 .... (1)\nand a(sub12) = a + 11d = 18 . (2)\nEquation (2) - equuation (1)\n7d = 28\nSo d = 4\nSo a = -26\na(sub 2) = a + d = -22\na(sub 17) = a + 16d = 38\n\n2. Find s(sub 4) for the geomtric sequence: \na(sub 1) = 5p, a(sub 2) = -10p\n\nLet a(sub 1) = a and common ratio be r\n\nFor a GP\na(sub n) = ar^(n-1)\n\na(sub 1) = a = 5p .......... (1)\na(sub 2) = ar =  -10p ..... (2)\nEquation (2)/equation (1)\nr = -2\n\nSo a(sub n) = 5p * (-2)^(n-1)\n\ns(sub n) = a(r^n - 1)/(r - 1)\n= 5p((-2)^n) - 1)/(-2 - 1)\n= -5p/3 ((-2)^n - 1)\n\nSo s(sub 4) = -5p/3 ((-2)^4 - 1)\n= -25p
272863
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.
426231
P(2 are women) = \nP(1st is woman) * P(2nd is woman) = \n6/10 * 5/9 = 30/90 = 0.333\n\nP(at least 1 is woman) =\n1 - P(both are men) =\n1 - P(1st is man)*P(2nd is man) = \n1 - (4/10)*(3/9) =\n1 - 12/90 =\n78/90 = 0.867\n\nI'm a little rusty on my "given that..." types of problems, and I'd hate to possibly steer you wrong on your remaining two questions.  But let me just say that I *THINK* the P of your 3rd question is 5/9; and I think the P of the 4th question is the same as the P of your 3rd question.\n\n~ ♥ ~
191098
Mutually exclusive means they can't both happen.\ncase 1, 2 is also even, not that one\ncase 2, 3 is also prime, not that one\ncase 3, an even number can't also be odd, that's the one\ncase 4, 4 is also >2, not that one\ncase 5, 1 is also <3, not that one
180437
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.
35666
F- even and odds.  There are 3 even sides and 3 odd sides. So the likelihood is the same for both.
404432
5 out of the 8 items in Box A are non-defective.\n3 out of the 5 items in Box B are non-defective.\n\nThe probability that an item from Box A and an item from Box B are drawn and are both non-defective is:\n\n5/8 * 3/5 = 3/8\n\nThe probability that an item from Box A and an item drawn from Box B and both defective is:\n\n3/8 * 2/5 = 3/20.\n\nThis means that the probability that one of the two items drawn is defective is:\n\n1 - p(both are not defective) - p(both are defective)\n=1 - 3/8 - 3/20\n= 19/40\n\nWe can see this by showing all possible outcomes. The items in Box A are shown on the top edge, and the items in Box B are shown on the right edge\n\nN  N  N  N  N  D  D D\nA  A  A   A  A  B  B  B  N\nA  A  A   A  A  B  B  B  N\nA  A  A   A  A  B  B  B  N\nB  B  B  B  B  C  C  C D\nB  B  B  B  B  C  C  C D\n\nThe events marked "A" represent 2 non-defective items. The events marked "B" represent 1 defective and 1 non-defective item. The events marked "C" represent 2 defective items.\n\nOut of 19 events where there is 1 defective and 1 non-defective item, the defective item came from box A 9 times.\n\nSo if 1 item is defective and 1 is not, the probability is 9/19 that it came from box A
562485
Find the probability of the complement occurring and then subtract that from 1.\nThe complement would be that an odd occurs 0 or 1 times.\nP(odd) = .5 and P(even) = .5\nP(0 odds) = C(10,0)(.5)^0 (.5)^10 = .0009765625\nP(1 odd) = C(10,1) (.5)^1 (.5)^9 = .009765625\n.0009765625 + .009765625 = .0107421875\n1 - .0107421875 = .9892578125
50512
A: 2 PPL HAVE SAME GENDER\nB: 2 PPL HAVE SAME BIRTHDAY\n\n\nP(A) = 90/190\nP(B) = 365/365 X 1/365 = 1/365\n\nP(AnB) = P(A) X P(B) = 90/190 X 1/365 = 0.001
444549
1) -2 x (2/5 -2p)\n    =4/5+4p            after multiplying through by -2\n\n2) 1 x (9/20 -p)\n    =9/20-p              after multiplying through by +1\n\nThis has started you off you should be able to solve now!
468276
P(B|Ac) = P(B and Ac)/P(Ac)\n\nP(B and Ac) = P(B) - P(A and B)  (It's easiest to see that if you look at a Venn diagram.  It's the stuff that both in B but not in A.)\n= 0.25-0.1\n= 0.15\n\nP(Ac) = 1-P(A) = 1-0.3 = 0.7\n\nSo P(B|Ac) = P(B and Ac)/P(Ac) = 0.15/0.7 = 15/70 = 3/14\n\nP(Bc|A) = 1 - P(B|A)\n= 1 - P(B and A)/P(A)\n= 1 - 0.1/0.3\n= 2/3\n\nAnswers would be similar for the next part.\n\nP(B|Ac) = 2/5\nP(Bc|A) = 17/20