Question

given ( p(a)=0.23 ) and ( p(b)=0.33 ), do the following.
a) if ( a ) and ( b ) are mutually exclusive events, compute ( p(a ) or ( b) )
b) if ( p(a ) and ( b)=0.1 ), compute ( p(a ) or ( b) )

Explanation:

Step1: Recall the formula for mutually - exclusive events

For mutually - exclusive events \(A\) and \(B\), \(P(A\cup B)=P(A)+P(B)\). Given \(P(A) = 0.23\) and \(P(B)=0.33\).

$$P(A\cup B)=0.23 + 0.33$$

Step2: Calculate the sum

$$0.23+0.33=0.56$$

Step3: Recall the general addition rule for non - mutually exclusive events

The general addition rule is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Given \(P(A) = 0.23\), \(P(B)=0.33\) and \(P(A\cap B)=0.1\).

$$P(A\cup B)=0.23 + 0.33-0.1$$

Step4: Calculate the result

$$0.23+0.33 - 0.1=0.46$$

Answer:

a) \(0.56\)
b) \(0.46\)