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the n partition of A , in B , so the results of summation of all Ai's probabilities which individually intersect with B divided by probability of B is totals theorem, so simply we say if you want to find the probability of any partition is bays theorem and if you have partitions and wants to find the probability of A is Totals theorem. (S.M SINDHI QUCEST LARKANA)
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Difference between conditional probability and Bayes theorem?
may b d reason is dat - in conditional prob. occurence of one of d 2 events is known 2 us . on other hand prob of d each event is given 2 us in baye's theorem.
Can you prove mathematically that a geometric random variable possesses the so called Memory-less property?
Proof: P{T>n+m/T>n}=P{T>n+m,T>n}/P{T>n} (Bayes theorem) =P{T>n+m}/P{T>n} =((1-p)^(n+m))/(1-p)^n = (1-p)^(n+m-n) = (1-p)^m (1-p)^m = {T>m} So T>m has the same probability as T>m+n given that T>n, which means it doesn't care (or don't remember) that n phases had passed.
How do you find p(b) when p(a) is 23 p(ba) is 12 and p(a U b) is 45 and is a dependent event?
There are symbols missing from your question which I cam struggling to guess and re-insert. p(a) = 2/3 p(b ??? a) = 1/2 p(a ∪ b) = 4/5 p(b) = ? Why use the set notation of Union on the third given probability whereas the second probability has something missing but the "sets" are in the other order, and the order wouldn't matter in sets. There are two possibilities: 1) The second probability is: p(b ∩ a) = p(a ∩ b) = 1/2 → p(a) + p(b) = p(a ∪ b) + p(a ∩ b) → p(b) = p(a ∪ b) + p(a ∩ b) - p(a) = 4/5 + 1/2 - 2/3 = 24/30 + 15/30 - 20/30 = 19/30 2) The second and third probabilities are probabilities of "given that", ie: p(b|a) = 1/2 p(a|b) = 4/5 → Use Bayes theorem: p(b)p(a|b) = p(a)p(b|a) → p(b) = (p(a)p(b|a))/p(a|b) = (2/3 × 1/2) / (4/5) = 2/3 × 1/2 × 5/4 = 5/12
Solution to a bag contains 4 balls Two balls are drawn at random and are found to be white What is the probability that all balls are white?
Two balls are certainly white. So the remaining balls may be one of the following{WW, WN, NW, NN}where W means a white ball and N means non-whiteballLet the events be:E- all four balls are white(WW)F- 3 balls are white(WN,NW)G- only two balls are white(NN)Now, P(E) =P(G) =1/4, and P(F) =1/2Let A be the event that the two balls drawn are whiteP(A|E)=4C2/4C2 =1P(A|F)=3C2/4C2 =1/2P(A|G)=2C2/4C2 =1/6Now By Bayes' Theorem,required probability =P(E|A)=(1*1/4) / ( 1*1/4 + 1/2*1/2 + 1/6*1/4 )=(1/4) / ( 13/24 )=6/13----------------------------------------------------------------------------------------------------2nd opinionLet's say we have 3 boxes with 4 balls each.Box A has 4 white balls.Box B has 3 white balls.Box C has 2 white balls.The probability of drawing 2 W balls from;Box A, P(2W│A) =(4/4)∙(3/3) =1Box B, P(2W│B) =(3/4)∙(2/3) =1/2Box C, P(2W│C) =(2/4)∙(1/3) =1/6Say the probability of picking any of the 3 boxes is the same, we have;P(A) =1/3P(B) =1/3P(C) =1/3Question is, given the event of drawing 2 W balls from a box taken blindlyfrom the 3 choices, what is the probability that the balls came from box A,P(A│2W).Recurring to Bayes Theorem:P(A│2W) =[P(A)P(2W│A)]/[P(A)P(2W│A)+P(B)P(2W│B)+P(C)P(2W│C)] =[(1/3)(1)]/[(1/3)(1)+(1/3)(1/2)+(1/3)(1/6)] =6/10 =0.60 =60%P(A│2W) =0.60 =60%
A bag contains 4balls two balls are drawn at random and are found to be white what is the probability that all balls are white?
the odds theoretically are almost infinity to 1 as they could be any color. -------------------------------------------------------------------------------------------2nd opinionLet's say we have 3 boxes with 4 balls each.Box A has 4 white balls.Box B has 3 white balls.Box C has 2 white balls.The probability of drawing 2 W balls from;Box A, P(2W│A)=(4/4)∙(3/3)=1Box B, P(2W│B)=(3/4)∙(2/3)=1/2Box C, P(2W│C)=(2/4)∙(1/3)=1/6Say the probability of picking any of the 3 boxes is the same, we have;P(A)=1/3P(B)=1/3P(C)=1/3Question is, given the event of drawing 2 W balls from a box taken blindlyfrom the 3 choices, what is the probability that the balls came from box A,P(A│2W).Recurring to Bayes Theorem:P(A│2W)=[P(A)P(2W│A)]/[P(A)P(2W│A)+P(B)P(2W│B)+P(C)P(2W│C)]=[(1/3)(1)]/[(1/3)(1)+(1/3)(1/2)+(1/3)(1/6)]=6/10=0.60=60%P(A│2W)=0.60=60%Read more:Solution_to_a_bag_contains_4_balls_Two_balls_are_drawn_at_random_and_are_found_to_be_white_What_is_the_probability_that_all_balls_are_white
