One satellite is scheduled to be launched from Cape Canaveral in Florida, and another launching is scheduled for Vandenberg Air Force Base in California. Let A denote the event that the Vandenberg launch goes off on schedule, and let B represent the event that the Cape Canaveral launch goes off on schedule. If A and B are independent events with P(A) > P(B), P(A ∪ B) 5 .626, and P(A ∩ B) 5 .144, determine the values of P(A) and P(B) to three decimal places.

Answer:[tex]P(A) = 0.45[/tex][tex]P(B) = 0.32[/tex]Step-by-step explanation:Given[tex]P(A) > P(B)[/tex][tex]P(A\ u\ B) = 0.626[/tex][tex]P(A\ n\ B) = 0.144[/tex]RequiredFind P(A) and P(B)We have that:[tex]P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)[/tex] --- (1)and[tex]P(A\ n\ B) = P(A) * P(B)[/tex] --- (2)The equations become:[tex]P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)[/tex] --- (1)[tex]0.626 = P(A) + P(B) - 0.144[/tex]Collect like terms[tex]P(A) + P(B) = 0.626 + 0.144[/tex][tex]P(A) + P(B) = 0.770[/tex]Make P(A) the subject[tex]P(A) = 0.770 - P(B)[/tex][tex]P(A\ n\ B) = P(A) * P(B)[/tex] --- (2)[tex]0.144 = P(A) * P(B)[/tex][tex]P(A) * P(B) = 0.144[/tex]Substitute: [tex]P(A) = 0.770 - P(B)[/tex][tex][0.770 - P(B)] * P(B) = 0.144[/tex]Open bracket[tex]0.770P(B) - P(B)^2 = 0.144[/tex]Represent P(B) with x[tex]0.770x - x^2 = 0.144[/tex]Rewrite as:[tex]x^2 - 0.770x + 0.144 = 0[/tex]Expand[tex]x^2 - 0.45x - 0.32x + 0.144 = 0[/tex]Factorize:[tex]x[x - 0.45] - 0.32[x - 0.45]= 0[/tex]Factor out x - 0.45[tex][x - 0.32][x - 0.45]= 0[/tex]Split[tex]x - 0.32= 0 \ or\ x - 0.45= 0[/tex]Solve for x[tex]x = 0.32\ or\ x = 0.45[/tex]Recall that:[tex]P(B) = x[/tex]So, we have:[tex]P(B) = 0.32 \ or \ P(B) = 0.45[/tex]Recall that:[tex]P(A) = 0.770 - P(B)[/tex]So, we have:[tex]P(A) = 0.770 - 0.32 \ or\ P(A) =0.770 - 0.45[/tex][tex]P(A) = 0.45 \ or\ P(A) =0.32[/tex]Since:[tex]P(A) > P(B)[/tex]Then:[tex]P(A) = 0.45[/tex][tex]P(B) = 0.32[/tex]