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## Thursday, 7 August 2014

### chapter-12 -worked-out-examples-5

 Example: 9
For two events $A$ and $B$, we are given the following information:
 $P(A) = \dfrac{1}{4},\;\;P(A/B) = \dfrac{1}{4}\;\;{\rm{and}}\;\;P(B/A) = \dfrac{1}{2}$
(a) Determine whether the two events are $ME$ or independent, or neither.
(b) Find $P(\bar B/\bar A)$. The notation $\bar A$ stands for the complementary event of event $A$, i.e., $\bar A$ occurs if $A$ does not occur. $(\bar A = not - A)$
 Solution: 9-(a)
Since $P(A)$ and $P(A/B)$ are the same, this means that $A$ and $B$ are independent events. Hence they are obviously not $ME$.
We therefore obtain
 $P(A \cap B) = P(A) \cdot P(B)$ (independence) $\Rightarrow \,\,\,\, P(B) = \dfrac{1}{2}$
 Solution: 9-(b)
Now, if $A$ and $B$ are independent events, then so are $\bar A$ and $\bar B$ (and in fact ($A$ and $\bar B$) and $(\bar A\;\,{\rm{and}}\,B)$. This fact should be intuitively obvious but let us justify it with some rigor. We have
 $\bar A \cap \bar B = (\overline {A \cup B} )$ $\left\{ \begin{array}{l} {\rm{From}}\,{\rm{elementary\, set \,theory; \,try\, to\, }}\\ {\rm{justify\, this\, using\, a \,venn\, diagram\,}} \end{array} \right\}$ $\Rightarrow \,\,\,\, P(\bar A \cap \bar B) = P(\overline {A \cup B} )$ $= 1 - P(A \cap B)$ $= 1 - \left\{ {P(A) + P(B) - P(A \cap B)} \right\}$ $= 1 - P(A) - P(B) + P(A) \cdot P(B)$ { since $A$, $B$ are independent } $= (1 - P(A))\;(1 - P(B))$ $= P(\bar A)\; \cdot \;P(\bar B)$ $\Rightarrow\,\,\,\, \bar A$ and $\bar B$ are independent
Similar proofs follow for the other two pairs.
Returning to the question, we see that since $\bar A$ and $\bar B$ are independent, we have
 $P(\bar B/\bar A) = P(\bar B) = 1 - P(B) = \dfrac{1}{2}$
 Example: 10
 In this example, we’ll get a taste of what is known as inverse probability, which we’ll of course be discussing in detail in a later section. A man is known to speak the truth $3$ out of $4$ times. He throws a die and tells you that he obtained a six. What is the chance of it being really a six ?
 Solution: 10
The problem here is that some event has already taken place, i.e., the throwing of a die. What we are required to do is find out how truthful the result being told to us is, given that the person reporting the result is not exactly a saint!
So, we first let $E$$F$$G$ denote the following events:
 $E$ : Man reports a six $F$ : Die shows a six $G$ : Die does not show a six
Observe that we are required to find $P(F/E)$, i.e., the probability of the die actually showing a six given that the man is reporting so.
Since we already know a bit about conditional probabilities, let us try to follow that route of manipulation.
 $P(F/E) = \dfrac{{P(F \cap E)}}{{P(E)}} = \dfrac{{P(F)P(E/F)}}{{P(E)}}$
Now, $F \cap E$ is the event that the man reports a six when actually he obtains a six. The probability of this event will simply be
 $P(F \cap E) = P(F)\; \cdot \;P(E/F)$ $= \dfrac{1}{6}\; \cdot \;\dfrac{3}{4}$ $= \dfrac{1}{8}$
You are urged to carefully understand how we arrived at this.
Next, we try to find $P(E)$, i.e., the probability of the man reporting a six. This can happen in two mutually exclusive ways.
 Way $P$(Way) Actual result : Six : Man says the truth $\dfrac{1}{6}\,\, \times \,\,\dfrac{3}{4}\,\, = \,\,\dfrac{1}{8}$ Actual result : Not a six : Man lies $\dfrac{5}{6}\,\, \times \,\,\dfrac{1}{4}\,\, = \,\,\dfrac{5}{{24}}$
Thus,
 $P(E) = \dfrac{1}{8} + \dfrac{5}{{24}}$ $= \dfrac{1}{3}$
Finally,
 $P(F/E) = \dfrac{{1/8}}{{1/3}}$ $= \dfrac{3}{8}$
It should be remarked that a lot of readers might not understand fully the underlying idea of this problem. They need not despair since inverse probabilities will be dealt with in much detail later.
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