Question

\and\ and \or\ events

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) = \frac{4}{13}=0.3076
c. find the event a or b.
p(a or b) =

Explanation:

Step1: Recall set - intersection formula for A and B

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 \(B=\{a,b,c,d,f,1,2,6\}\), we find the common elements. So \(A\cap B=\{b,c,d,6\}\).

Step2: Recall probability formula for \(P(A\cap B)\)

The probability of an event \(E\) in a sample - space \(S\) is given by \(P(E)=\frac{n(E)}{n(S)}\), where \(n(E)\) is the number of elements in event \(E\) and \(n(S)\) is the number of elements in the sample - space. 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 for A or B

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). \(A\cup B=\{a,b,c,d,f,g,1,2,4,6\}\).

Answer:

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