Question

ch 12 a fair die is rolled and the sample space is given s={1,2,3,4,5,6}. let a={1,2} and b={3,4}. which of the following statements is true? p(a or b)=4/6. p(a or b) = p(a)+p(b)-p(a and b). a and b are mutually exclusive (no elements in common). p(a or b) = p(a)+p(b).

Explanation:

Step1: Recall probability formula for mutually - exclusive events

For two mutually - exclusive events \(A\) and \(B\) (events with no common elements), the probability of \(A\) or \(B\) is given by \(P(A\cup B)=P(A)+P(B)\). The probability of an event \(E\) in a sample space \(S\) with \(n(S)\) elements is \(P(E)=\frac{n(E)}{n(S)}\). Here, \(n(S) = 6\), \(n(A)=2\), \(n(B)=2\).

Step2: Calculate \(P(A)\) and \(P(B)\)

\(P(A)=\frac{n(A)}{n(S)}=\frac{2}{6}\), \(P(B)=\frac{n(B)}{n(S)}=\frac{2}{6}\).

Step3: Calculate \(P(A\cup B)\)

Since \(A\cap B=\varnothing\) (no common elements, mutually - exclusive), \(P(A\cup B)=P(A)+P(B)=\frac{2}{6}+\frac{2}{6}=\frac{4}{6}\). Also, the general formula for \(P(A\cup B)\) is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), and when \(A\) and \(B\) are mutually - exclusive \(P(A\cap B) = 0\).

Answer:

All of the statements are true.

1. \(P(A\ or\ B)=\frac{4}{6}\) because \(P(A)=\frac{2}{6}\), \(P(B)=\frac{2}{6}\) and \(A\) and \(B\) are mutually exclusive so \(P(A\ or\ B)=P(A)+P(B)=\frac{2 + 2}{6}=\frac{4}{6}\).
2. \(P(A\ or\ B)=P(A)+P(B)-P(A\ and\ B)\) is the general addition rule for probability.
3. \(A=\{1,2\}\) and \(B = \{3,4\}\) have no elements in common, so \(A\) and \(B\) are mutually exclusive.
4. Since \(A\) and \(B\) are mutually exclusive (\(P(A\ and\ B)=0\)), \(P(A\ or\ B)=P(A)+P(B)\).