Mathematics: 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 )

Let S be a sample space and E and F be events associated

oomale

30.01.2022

(a)0.5

(b)0.17

(b)0.375

Step-by-step explanation:

Pr(E)=0.6

Pr(F)=0.2

$Pr(E\cap F)=0.1.$

(a)Pr (E|F )

$Pr (E|F )=\dfrac{Pr(E \cap F)}{Pr(F)} \\=\dfrac{0.1}{0.2}\\\\=0.5$

(b)Pr (F|E )

$Pr (F|E )=\dfrac{Pr(E \cap F)}{Pr(E)} \\=\dfrac{0.1}{0.6}\\\\=0.17$

(c)Pr (E|F')

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

=1-0.2=0.8

$Pr(E \cap F')=P(E)-P(E\cap F)\\=0.6-0.1\\=0.5$

Therefore:

$Pr (E|F' )=\dfrac{Pr(E \cap F')}{Pr(F')} \\=\dfrac{0.5}{0.8}\\\\=0.625$

(d)Pr(E'|F')

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

Therefore:

$Pr (E'|F' )=\dfrac{Pr(E' \cap F')}{Pr(F')} \\=\dfrac{0.3}{0.8}\\\\=0.375$

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yeontan

15.05.2021

42.29

Step-by-step explanation:

First, multiply 1/4 (or 0.25) by the area of the pizza:

56.52 x 0.25 = 14.13

Then subtract that value from the whole area:

56.52 - 14.13 = 42.29

42.29 is your answer

Hope this helps!

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