find the indicated probability\ngiven\np(a) = 0.4\np(b) = 0.2\np(a or b) = 0.6\nfind\np(a and b) =\nare…
find the indicated probability\ngiven\np(a) = 0.4\np(b) = 0.2\np(a or b) = 0.6\nfind\np(a and b) =\nare these events disjoint?\nyes\nno
Answer
Explanation:
Step1: Recall the addition - rule formula
The formula for $P(A\ or\ B)$ is $P(A\ or\ B)=P(A)+P(B)-P(A\ and\ B)$.
Step2: Rearrange the formula to solve for $P(A\ and\ B)$
We can rewrite the formula as $P(A\ and\ B)=P(A)+P(B)-P(A\ or\ B)$.
Step3: Substitute the given values
Substitute $P(A) = 0.4$, $P(B)=0.2$ and $P(A\ or\ B)=0.6$ into the formula. So, $P(A\ and\ B)=0.4 + 0.2-0.6$.
Step4: Calculate the value of $P(A\ and\ B)$
$P(A\ and\ B)=0$.
Step5: Determine if the events are disjoint
Two events $A$ and $B$ are disjoint if $P(A\ and\ B) = 0$. Since $P(A\ and\ B)=0$, the events are disjoint.
Answer:
$P(A\ and\ B)=0$ Yes