Let E1,.., En be independent events. Prove that P(E1 U..U En) = 1 – 1[[1 – P(E:)]. i=1
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- The conditional probability of E given that F occurs is P(EF)=___________. So in rolling a die the conditional probability of the event E, getting a six, given that the event F, getting an even number, has occurred is P(EF)=___________.Prove that P(A') = 1 − P(A), for any event A.1. If there are n random variables such that each expectation exist then prove that E(X1+…….+Xn ) = E(X1)+……..+E(Xn). Assume discrete.
- Let X and Y are independent Poisson random variables such that E(X) = E(Y)=2. Let Z=X+Y. Compute P(X=2|Z=3).Assume that E and F are independent events. If PR (E) = 0.7 and Pr (F|E) = 0.6, find Pr(F)Let X1 and X2 be independent random variables; X1 ∼ Poisson(2); X2 ∼ Poisson(3). Compute E(X1|X1 +X2 = 10).
- If E and F are two independent events, such that P (E ∩ F ) = 1/6 , P (E' ∩ F' ) = 1/3 and (P (E) − P (F ))(1 − P (F )) > 0, then derive the relation between P(E) and P(F).Let E and F be two events in S with P(E) = 0.54, P(F) = 0.57, and P(E ∩ F) = 0.21. Find P(EC ∩ F). (Enter answer as a decimal with at least 2 correct decimal places)Let E1 and E2 be events with P(E1)>0 and P(E2)>0. Prove that if P(E1|E2)>P(E1) then P(E2|E1)>P(E2).
- The events E, F and G have the following properties: P(E)=0.32 P(F)=0.39 P(G)=0.28P(E)=0.32 P(F)=0.39 P(G)=0.28 P(EUF)=P(EUF)=0.60 P(EUG)=0.45 P(FUG)=0.44P(EUG)=0.45 P(FUG)=0.44 P(E∩F∩G)=P(E∩F∩G)=0.04 Compute the probability that at least one of E,FE,F or GG occur; that is, compute P(EUFUG)P(EUFUG). Use two decimal place accuracy.Let A and B be two events such that 0 < P(A) < 1. If P(B|A) = P(B|Ac), prove mathematically(not intuitively) that events A and B are independent. Show all the steps of the mathematicalproof.Let X1, X2, X3 be random variables such that V ar(X1) = 5, V ar(X2) = 4,V ar(X3) = 7, cov(X1, X2) = 3, cov(X1, X3) = −2 and X2 and X3 are independent. Findthe covariance between Y1 = X1 − 2X2 + 3X3 and Y2 = −2X1 + 3X2 + 4X3