Question

Using the following information = Universal set = {2,3,4,5,6,7,8,9,12,13,14,16,20,22,56). Subset A = {9,12,13,20,22,56); Subset H= {4,5,8,9,16,22) and Subset C = {1,4,20,22,56} Find: a) P [(AnH)UC] (b) P (HnC)’ (c) P(H)’ (d) P[C\H] (e) P [A\C]

Answer 1

$\dfrac{2}{5},\dfrac{13}{15},\dfrac{3}{5},\dfrac{1}{5},\dfrac{1}{5}$

step-by-step explanation:

The intersection of two sets A and B, denoted by A ∩ B, is the set containing all elements of A that also belong to B.

The union (denoted by ∪) of a collection of sets is the set of all elements in the collection.

$P(s)$ of a set $s$ is defined as ratio of number of elements in $s$ to the number of elements in $universal set$

given $Universalset=$$\{$$\text{2,3,4,5,6,7,8,9,12,13,14,16,20,22,56}$$\}$

given $A=\{9,12,13,20,22,56\}$ and $H=\{4,5,8,9,16,22\}$

and $C=\{1,4,20,22,56\}$

For Question A:

$A$∩$H$=$\{9,12,13,20,22,56\}$ ∩ $\{4,5,8,9,16,22\}$

=$\{9,22}\}$

($A$∩$H$)∪$C$=$\{9,22}\}$ ∪ $C=\{1,4,20,22,56\}$=$\{1,4,9,20,22,56\}$

$p((A$∩$H)$∪$C)$=$\frac{6}{15}=\frac{2}{5}$

For Question B:

$H$∩$C$=$\{4,22\}$

$p(H$∩$C)$'=$\frac{15-2}{15}=\frac{13}{15}$

For Question C:

$p(H)$'=$\frac{15-6}{15}=\frac{3}{5}$

For Question D:

$C$\$H$=$\{1,20,56\}$

$p(C$\$H)$=$\frac{3}{15}=\frac{1}{5}$

For Question E:

$A$\$C$=$\{9,12,13\}$

$p(A$\$C)$=$\frac{3}{15}=\frac{1}{5}$