Se for usada a urna A, a probabilidade de retirar 2 bolas pretas é __ . __ = 10 9 2 1__ = __. 90 45 Se for usada a urna B, a probabilidade de reti...

Para resolver esse problema, precisamos calcular as probabilidades de retirar 2 bolas pretas de cada urna e, em seguida, a probabilidade de ter sido usada a urna A, dado que foram retiradas 2 bolas pretas. Para a urna A: A probabilidade de retirar 2 bolas pretas é (10/90) * (9/89) = 1/89. Para a urna B: A probabilidade de retirar 2 bolas pretas é (56/90) * (28/89) = 14/45. Agora, para calcular a probabilidade de ter sido usada a urna A, dado que foram retiradas 2 bolas pretas, usamos a fórmula de probabilidade condicional: P(A|B) = P(A ∩ B) / P(B), onde P(A ∩ B) é a probabilidade de A e B ocorrerem e P(B) é a probabilidade de B ocorrer. Assim, a probabilidade de ter sido usada a urna A, dado que foram retiradas 2 bolas pretas, é: (1/89) / ((1/89) + (14/45)) = (1/89) / (89/89 + 28/45) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / (89/89 + 56/90) = (1/89) / 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