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: Calculate probabilities of A and B

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

Step2: Check mutual - exclusivity

$A=\{1,2\}$, $B = \{3,4\}$, $A\cap B=\varnothing$, so A and B are mutually exclusive.

Step3: Apply addition rule for mutually - exclusive events

For mutually - exclusive events, $P(A\cup B)=P(A)+P(B)$. Since $P(A)=\frac{2}{6}$ and $P(B)=\frac{2}{6}$, $P(A\cup B)=\frac{2 + 2}{6}=\frac{4}{6}$. Also, the general addition rule is $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. When $A$ and $B$ are mutually exclusive, $P(A\cap B) = 0$, so $P(A\cup B)=P(A)+P(B)$.

Answer:

All of the statements are true.