Let S be a sample space and E and F be events associated with S. Suppose that Pr (Upper E )equals 0.6, Pr (Upper F )equals 0.2 and Pr (Upper E intersect Upper F )equals 0.1. Calculate the following probabilities. (a) Pr (E|F )(b) Pr (F|E )(c) Pr (E| Upper F prime )(d) Pr (Upper E prime | Upper F prime )

Question

contestada

Respuesta :

Newton9022 Newton9022 · Answer 1 · 2020-07-13T02:16:15+02:00

Answer:

(a)0.5

(b)0.17

(c)0.625

(b)0.375

Step-by-step explanation:

Pr(E)=0.6

Pr(F)=0.2

[tex]Pr(E\cap F)=0.1.[/tex]

(a)Pr (E|F )

[tex]Pr (E|F )=\dfrac{Pr(E \cap F)}{Pr(F)} \\=\dfrac{0.1}{0.2}\\\\=0.5[/tex]

(b)Pr (F|E )

[tex]Pr (F|E )=\dfrac{Pr(E \cap F)}{Pr(E)} \\=\dfrac{0.1}{0.6}\\\\=0.17[/tex]

(c)Pr (E|F')

Pr(F')=1-P(F)

=1-0.2=0.8

[tex]Pr(E \cap F')=P(E)-P(E\cap F)\\=0.6-0.1\\=0.5[/tex]

Therefore:

[tex]Pr (E|F' )=\dfrac{Pr(E \cap F')}{Pr(F')} \\=\dfrac{0.5}{0.8}\\\\=0.625[/tex]

(d)Pr(E'|F')

[tex]P(E'\cap F')=P(E \cup F)'\\=1-P(E \cup F)\\=1-[P(E)+P(F)-P(E\cap F)]\\=1-[0.6+0.2-0.1]\\=1-0.7\\=0.3[/tex]

Therefore:

[tex]Pr (E'|F' )=\dfrac{Pr(E' \cap F')}{Pr(F')} \\=\dfrac{0.3}{0.8}\\\\=0.375[/tex]