Printable View
The events A, B and C are such that A and B are independent, and A and C are mutually exclusive. Given that P(A)=0.4, P(B)=0.2, P(C)=0.3, P(B ∩C)=0.1, calculate P(B/A∪C). Also, calculate the probability that one and only one of the events B, C will occur
Hello, Punch!
Quote:
The events $\displaystyle A, B, C$ are such that $\displaystyle A$ and $\displaystyle B$ are independent,
and $\displaystyle A$ and $\displaystyle C$ are mutually exclusive.
Given:.$\displaystyle P(A)=0.4\;\;\;P(B)=0.2\;\;\;P(C)=0.3\;\;\;P(B \cap C)=0.1$
Draw a Venn diagram.
$\displaystyle A$ and $\displaystyle C$ are mutually exclusive.Code:* * * * * * * * *
* A * B * C *
* * * * * *
* * * * * *
* * * * * *
* 0.32 * 0.08 * * 0.1 * 0.2 *
* * * * * *
* * * * * *
* * * 0.02 * * *
* * * *
* * * * * * * * *
. . Their circles do not intersect.
$\displaystyle A$ and $\displaystyle B$ are independent.
Hence:.$\displaystyle P(A \cap B) \:=\:P(A)\cdot P(B) \:=\:(0.4)(0.2) \:=\:0.08$
$\displaystyle P(B \cap C) \,=\,0.1$
We can complete the Venn diagram.
Quote:
(a) Calculate $\displaystyle P(B\,|\,A \cup C)$
Bayes' Theorem: .$\displaystyle P(B\,|\,A\cup C) \;=\;\frac{P(B \cap [A\cup C])}{P(A\cup C)} $
From the Venn diagram, we see that:
. . $\displaystyle P(B\cap[A \cup C]) \:=\:0.08 + 0.1 \:=\:0.18$
. . $\displaystyle P(A \cup C) \:=\:0.4 + 0.3 \:=\:0.7$
$\displaystyle \text{Therefore: }\:P(B\,|\,A\cup C) \:=\:\frac{0.18}{0.70} \:=\:\frac{9}{35}$
Quote:
(b) Calculate the probability that exactly one one of $\displaystyle B, C$ will occur.
From the Venn diagram, we see that:
. . $\displaystyle P(B\text{ only}) \:=\:0.1$
. . $\displaystyle P(C\text{ only}) \:=\:0.2$
Therefore:.$\displaystyle P(B\text{ only}\,\cup\, C\text{ only}) \:=\:0.1 + 0.2 \:=\:0.3$
Thank you very much! Venn diagrams are certain useful in solving the question, but I have a doubt:
Based on the rule for independent events, $\displaystyle P(A\capB)=P(A)P(B)$
So, why is this not true: $\displaystyle P(B|A\cupC)=\frac{P(B\cap[A\cup C])}{P(A\cup C)}=\frac{P(B)P(A\cup C)}{P(A\cup C)}=P(B)$
