The National Association of State Contractors Licensing Agencies (NASCLA) is committed to providing quality resources to its members! One of the ways the association strives to accomplish this is by publishing a quarterly Newsletter. NASCLA’s quarterly newsletter keeps you up to date on the association meetings and projects, articles highlighting current events in the construction industry, as well as a Spotlight section featuring NASCLA members.
The National Association of State Contractors Licensing Agencies (NASCLA) is committed to providing quality resources to its members! One of the ways the association strives to accomplish this is by publishing a quarterly Newsletter. NASCLA’s quarterly newsletter keeps you up to date on the association meetings and projects, articles highlighting current events in the construction industry, as well as a Spotlight section featuring NASCLA members.
2. En una Unidad de Enfermería hay
110 enfermeras. De ellas 52
participan en la investigación I, 36 en
la investigación II. De esas 52 y 36
enfermeras, 12 en las dos
investigaciones.
Se elige al azar una enfermera.
Determinar las probabilidades de los
siguientes sucesos:
4. • 1. Participa en la investigación I:
P(I) = CF/CP = 52/110 = 0.473
• 2. Participa en II:
P(II) = CF/CP = 36/110 = 0.327
• 3. Solo participa en II:
P(solo II) = CF/CP = 24/110 = 0.218
• 4. Participa, a la vez, en I y II:
P(I Y II) = P(IῼII) = CF/CP = 12/110 = 0.109
5. • 5. Participa en I o en II:
P(I U II) = P(I) + P(II) – P(I ῼ II) =
= 0.473 + 0.327 – 0.109 = 0.691
• 6. Participa solo en una investigación:
P(solo en una investigación)= 40+24/110= 0.581
• 7. No participa en alguna investigación:
P (ni en I ni II) = CF/CP = 22/110 = 0.2
6. • 8. No participa en I:
P (no I) = 1 – P(I) = 1 – 0.473 = 0.527
• 9. Participa en II sabiendo que participa en I:
P(II ǀ I) = P(II ῼ I)/P(I) = 0.109/0.473 = 0.230
• 10. Participa en I sabiendo que participa en II:
P(I ǀ II) = P(I ῼ II)/P(II) = 0.109/0.327 = 0.333