1 like 0 dislike
198 views
Two events $\mathrm{A}$ and $\mathrm{B}$ are such that $\mathrm{P}(A)=0.5, \mathrm{P}(B)=0.3$ and $\mathrm{P}(A \cap B)=0.1 .$

Calculate
(a) $\mathrm{P}(A \mid B)$;
(b) $\mathrm{P}(B \mid A)$;
(c) $\mathrm{P}(A \mid A \cup B)$;
(d) $\mathrm{P}(A \mid A \cap B)$;
(e) $\mathrm{P}(A \cap B \mid A \cup B)$.
| 198 views

0 like 0 dislike
(Venn diagrams are helpful in understanding some of the events that arise below.)
(a) $\mathrm{P}(A \mid B)=\mathrm{P}(A \cap B) / \mathrm{P}(B)=\frac{1}{3}$

(b) $\mathrm{P}(B \mid A)=\mathrm{P}(A \cap B) / \mathrm{P}(A)=\frac{1}{5}$

(c) $\mathrm{P}(A \cup B)=\mathrm{P}(A)+\mathrm{P}(B)-\mathrm{P}(A \cap B)=0.7$, and the event $A \cap(A \cup B)=A$, so
$$\mathrm{P}(A \mid A \cup B)=\mathrm{P}(A) / \mathrm{P}(A \cup B)=\frac{5}{7}$$

(d) $\mathrm{P}(A \mid A \cap B)=\mathrm{P}(A \cap B) / \mathrm{P}(A \cap B)=1$, since $A \cap(A \cap B)=A \cap B$.

(e) $\mathrm{P}(A \cap B \mid A \cup B)=\mathrm{P}(A \cap B) / \mathrm{P}(A \cup B)=\frac{1}{7}$, since $A \cap B \cap(A \cup B)=A \cap B$.
by Platinum (131,394 points)

2 like 0 dislike
1 like 0 dislike
0 like 0 dislike
1 like 0 dislike
2 like 0 dislike
0 like 0 dislike
1 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
1 like 0 dislike
1 like 0 dislike