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 EFG 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.
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}}
P(E) = \dfrac{1}{8} + \dfrac{5}{{24}}
 = \dfrac{1}{3}
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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