(a) Let P(B1∩B2)>0, and A1∪A2⊂B1∩B2. Then show that
P(A1|B1).P(A2|B2)=P(A1|B2).P(A2|B1).
(b) Let A and B1 be independent; similarly, let A and B2 be independent. Show that in this case, A and B1∪B2 are independent if and only if A and B1∩B2 are independent.
(c) Given P(A) = 0.42,P(B) = 0.25, and P(A∩B) = 0.17, find
(i)P(A∪B) ;
(ii)P(A∩Bc) ;
(iii)P(Ac∩Bc) ;
(iv)P(Ac|Bc).
Complete solution is given in attached images:
Thank You.
(a) Let P(B1∩B2)>0, and A1∪A2⊂B1∩B2. Then show that P(A1|B1).P(A2|B2)=P(A1|B2).P(A2|B1). (b) Let A and B1 be independent;...
2. (a) Let P(Bin B2) > 0, and AUA, CBin B2. Then show that P(A/B).P (A2|B2) = P(A|B2).P (A2|Bi). (b) Let A and Bbe independent; similarly, let A and B, be independent. Show that in this case, A and B U B2 are independent if and only if A and Bin B2 are independent (c) Given P(A) = 0.42, P(B) = 0.25, and P(An B) = 0.17, find (i) P (AUB); (ii) P(An B°); (iii) P(A n B); (iv) P(...
Show that the formula for a line through two points (a1,b1) and (a2,b2) is: y=(b1-b2)/(a1-a2) * x + (a1*b2-a2*b1)/(a1-a2) The slope part looks right. Could someone explain how the y-intercept part makes sense? I would think that the y-intercept is: b1-slope*a1 or b2-slope*a2 anyone have any ideas?
GAME 3 Player B B1 B2 Player A A1 7,3 5, 10 A2 3,8 9,6 In Game 3 above, if the players move sequentially with Player B choosing first, the Nash equilibrium will be a) Player A choosing A2 and Player B choosing B1 b) Player A choosing A2 and Player B choosing B2 c) Player A choosing A1 and Player B choosing B2 d) Player A choosing A1 and Player B choosing B1
Given these probabilities, complete the contingency table, and compute the following probabilities: a) P(A2 and B1) b) P(A1 | B1) c) P(B2 | A2) d) P(B2 or A1) A1 A2 Total B1 0.56 B2 Total 0.46
given the following joint probability table A1 A2 B1 .02 .01 B2 .05 .02 Calculate the conditional probability P(A1IB1) round your answer
float useless(A){ n = A.length; if (n==1) { return A[@]; let A1,A2 be arrays of size n/2 for (i=0; i <= (n/2)-1; i++){ A1[i] = A[i]; A2[i] = A[n/2 + i]; for (i=0; i<=(n/2)-1; i++){ for (j=i+1; j<= (n/2)-1; j++){ if (A1[i] == A2[j]) A2[j] = 0; b1 = useless(A1); b2 = useless (A2); return max(b1,b2); What is the asymptotic upper bound of the code above?
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