Question

1. let the sample space, s = {a,b,c,d,e,f,g,1,2,3,4,5,6}, event a = {b,c,d,g,4,6}, and event b = {a,b,c,d,f,1,2,6}.

a. find the event a and b.

a and b = {b,c,d,6}

b. what is the probability of obtaining event a and b?

p(a and b) = 4/13 = 0.3076

c. find the event a or b.

a or b = {a,b,c,d,f,g,1,2,4,6}

d. what is the probability of obtaining event a or b?

p(a or b) = 10/13 = 0.7692

e. what is the complement of event a?

Explanation:

Step1: Recall set - intersection formula

The intersection of two sets \(A\) and \(B\) (denoted as \(A\cap B\)) is the set of all elements that are in both \(A\) and \(B\). Given \(A = \{b,c,d,g,4,6\}\) and \(B=\{a,b,c,d,f,1,2,6\}\), \(A\cap B=\{b,c,d,6\}\).

Step2: Calculate probability of \(A\cap B\)

The probability \(P(A\cap B)=\frac{n(A\cap B)}{n(S)}\), where \(n(A\cap B)\) is the number of elements in \(A\cap B\) and \(n(S)\) is the number of elements in the sample - space \(S\). Here, \(n(A\cap B) = 4\) and \(n(S)=13\), so \(P(A\cap B)=\frac{4}{13}\approx0.3077\).

Step3: Recall set - union formula

The union of two sets \(A\) and \(B\) (denoted as \(A\cup B\)) is the set of all elements that are in \(A\) or \(B\) (or both). \(A\cup B=\{a,b,c,d,f,g,1,2,4,6\}\).

Step4: Calculate probability of \(A\cup B\)

The probability \(P(A\cup B)=\frac{n(A\cup B)}{n(S)}\). Here, \(n(A\cup B) = 10\) and \(n(S)=13\), so \(P(A\cup B)=\frac{10}{13}\approx0.7692\).

Step5: Recall complement formula

The complement of set \(A\) (denoted as \(A^c\)) is the set of all elements in the sample - space \(S\) that are not in \(A\). So \(A^c=\{a,f,1,2,3,5\}\).

Answer:

a. \(A\cap B = \{b,c,d,6\}\)
b. \(P(A\cap B)=\frac{4}{13}\approx0.3077\)
c. \(A\cup B=\{a,b,c,d,f,g,1,2,4,6\}\)
d. \(P(A\cup B)=\frac{10}{13}\approx0.7692\)
e. \(A^c=\{a,f,1,2,3,5\}\)