Math Help - Probability using set notation

1. Probability using set notation

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/AC). Also, calculate the probability that one and only one of the events B, C will occur

2. Re: Probability using set notation

Hello, Punch!

The events $A, B, C$ are such that $A$ and $B$ are independent,
and $A$ and $C$ are mutually exclusive.

Given:. $P(A)=0.4\;\;\;P(B)=0.2\;\;\;P(C)=0.3\;\;\;P(B \cap C)=0.1$

Draw a Venn diagram.
Code:
              * * *       * * *       * * *
*     A     *     B     *     C     *
*           *   *       *   *           *
*           *     *     *     *           *

*           *       *   *       *           *
*   0.32    *  0.08 *   *  0.1  *    0.2    *
*           *       *   *       *           *

*           *     *     *     *           *
*           *   *  0.02 *   *           *
*           *           *           *
* * *       * * *       * * *
$A$ and $C$ are mutually exclusive.
. . Their circles do not intersect.

$A$ and $B$ are independent.
Hence:. $P(A \cap B) \:=\:P(A)\cdot P(B) \:=\:(0.4)(0.2) \:=\:0.08$

$P(B \cap C) \,=\,0.1$

We can complete the Venn diagram.

(a) Calculate $P(B\,|\,A \cup C)$

Bayes' Theorem: . $P(B\,|\,A\cup C) \;=\;\frac{P(B \cap [A\cup C])}{P(A\cup C)}$

From the Venn diagram, we see that:
. . $P(B\cap[A \cup C]) \:=\:0.08 + 0.1 \:=\:0.18$
. . $P(A \cup C) \:=\:0.4 + 0.3 \:=\:0.7$

$\text{Therefore: }\:P(B\,|\,A\cup C) \:=\:\frac{0.18}{0.70} \:=\:\frac{9}{35}$

(b) Calculate the probability that exactly one one of $B, C$ will occur.

From the Venn diagram, we see that:
. . $P(B\text{ only}) \:=\:0.1$
. . $P(C\text{ only}) \:=\:0.2$

Therefore:. $P(B\text{ only}\,\cup\, C\text{ only}) \:=\:0.1 + 0.2 \:=\:0.3$

3. Re: Probability using set notation

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, $P(A\capB)=P(A)P(B)$

So, why is this not true: $P(B|A\cupC)=\frac{P(B\cap[A\cup C])}{P(A\cup C)}=\frac{P(B)P(A\cup C)}{P(A\cup C)}=P(B)$