tgt

## Thursday, 7 August 2014

### chapter-5 Example of Drawing a Card – II

In the experiment of drawing one card at random from a standard well-shuffled deck of $52$ cards, consider the event $E$ defined as:
 $E$ : The card drawn is a king
Obviously, we have
 $P\left( E \right) = \dfrac{{\# {\rm{Kings}}}}{{\# {\rm{All}}\,{\rm{cards}}}}$ $= \dfrac{4}{{52}}$ $= \dfrac{1}{{13}}$
However, if you are now asked to find the probability of event $E$ occurring given that the card drawn has a number greater than $10$, what would you say?
To answer correctly, you must understand that now the situation is entirely different from earlier. Now we already know that the card drawn has a number greater than $10$, which means that the number of possibilities for the card has reduced; earlier, there were $52$ possibilities, but now the number of possible cards are the $4$ Jacks, the $4$ Queens and the $4$ Kings, which means that now there are only $12$ possibilities for the card. Of these, the favorable possibilities are $4$ ; thus,
 $P\left\{ \begin{array}{l} {\rm{Event }}\,E\,{\rm{ given\, that \,the \,card\, drawn }}\\ {\rm{has\, a\, number\, greater\, that \,10\,}} \end{array} \right\} = \dfrac{{\# \,\,{\rm{Kings}}}} {{{\rm{\# \,Reduced\, possibilities}}}}$ $= \dfrac{4}{{12}}$ $= \dfrac{1}{3}$
Let us denote by $F$ the event that the card drawn has a number greater than $10$. The probability just calculated above is then written in standard notation as
 $P\left( {E/F} \right)$
 $P\left( {E\,\,{\rm{given}}\,\,F} \right)$
i.e. “the probability of event $E$ occurring given that $F$ has occurred”.
Let us consider another example. In the random experiment of throwing two dice, let events $G$ and $H$ be defined as
 $G$ : The sum of the two numbers on top is $6$ $H$ : One of the numbers on top is $4$
First, let us find the probability of $G$ occurring. This is simply
 $P\left( G \right) = \dfrac{{\# {\rm{favorable}}\,\,{\rm{possibilities}}}}{{\# \,{\rm{total}}\,\,{\rm{possibilites}}}}$ $= \dfrac{{\left( {1,5} \right)\,\,\left( {2,4} \right)\,\,\left( {3,3} \right)\,\,\left( {4,2} \right)\,\,\left( {5,1} \right)}}{{6 \times 6}}$ $= \dfrac{5}{{36}}$
Now, suppose we are given that $H$ has already occurred, meaning we now know that one of the numbers on top is $4$. What is the probability of $G$ now when we already possess this information?
The number of total possibilities now are $11$; we list them explicitly:
 $\left\{ {\begin{array}{*{20}{c}} {\left( {4,1} \right)}&{\left( {4,2} \right)}&{\left( {4,3} \right)}&{\left( {4,4} \right)}&{\left( {4,5} \right)}&{\left( {4,6} \right)}\\ {\left( {1,4} \right)}&{\left( {2,4} \right)}&{\left( {3,4} \right)}&{}&{\left( {5,4} \right)}&{\left( {6,4} \right)} \end{array}} \right\}$
Out of these, the favorable possibilities are only $2$, namely: $(4, 2)$ and $(2, 4)$. Thus
 $P\left( {G/H} \right) = \dfrac{{\# {\rm{favorable}}\,\,{\rm{possibilities}}}}{{\# {\rm{reduced \,possibilities}}}}$ $=\dfrac{2}{11}$
In both the preceding examples, what happens is that with the possession of some information, the number of possibilities reduce, i.e., the sample space reduces. In the first example, with the information that $F$ has occurred, the number of possibilities reduces from $52$ to $12$. We then looked for favorable cases only within this reduced sample space of $12$ outcomes. Similarly, in the second example, when we possess the information that $H$ has occurred, the number of possibilities reduces from $36$ to $11$; it is in this reduced sample space of outcomes that we then looked for the favorable possibilities.
In passing we should mention that probabilities of the form $P(A/B)$ are called conditional probabilities. $P(A/B)$ is said to be the conditional probability of $A$ given that $B$ has occurred.
Let us understand all this somewhat more deeply. In particular, let us ponder over the following question: Let $E$ and $F$ be two events. Let the probability of $E$ occurring be $P(E)$. Now, if we come to know that $F$ has occurred, will the probability of $E$ occurring always increase, like it did in the last two examples, or can it decrease too? Can it remain the same?
It turns out that all the three are possible, which should even be obvious to an alert reader.
Let us consider the case where the information that $F$ has occurred decreases the probability of $E$ occurring. Let $E$ and $F$ be, in the card drawing experiment of the first example
 $E$ : The card drawn has a number which is a multiple of four $F$ : The card drawn has a number greater than eight
The original probability of $E$ occuring is
 $P\left( E \right) = \dfrac{{\# {\rm{favorable}}\,\,{\rm{possibilities}}}}{{\# \,{\rm{total}}\,\,{\rm{possibilites}}}}$ $= \dfrac{{\left\{ {4,8,12} \right\}{\rm{per\, suit}}\,\, \times \,\,{\rm{4}}\,\,{\rm{suits}}}}{{52}}$ $= \dfrac{{12}}{{52}}\,\, = \,\,\dfrac{3}{{13}}$
But when $F$ has already occurred, the probability of $E$ occurring is
 $P\left( {E/F} \right) = \dfrac{{\# {\rm{favorable}}\,\,{\rm{possibilities}}}}{{\# \,\,{\rm{reduced}}\,\,{\rm{possibilites}}}}$ $= \dfrac{{4\,\,{\rm{Queens}}}}{{\left\{ {9,10,J,Q,K} \right\}\,{\rm{ per \,suit }} \times \,\,\,4\,{\rm{suits}}}}$ $= \dfrac{4}{{{\rm{20}}}} = \dfrac{1}{5}$
which is lesser than the original probability of $E$ occurring. This should be intuitively obvious: In the first case, there are $3$ multiples of four per suit of $13$cards. In the second, when we are told that the number of the card is greater than eight, the multiple of four possible is only $1$ per suit which is the queen. Thus the favorable possibilities decrease to one-third. The total possibilities also decrease i.e, from $13$ to $5$ per suit, but it can be easily appreciated that the percentage reduction in favorable possibilities is greater than the percentage reduction in total possibilities.
We now come to the third very important question: for two events $E$ and $F$, can $P(E/F)$ be the same as $P(E)$? You must appreciate that this is equivalent to saying that the probability of $E$ occurring is not affected by the occurrence or non-occurrence of $F$.
As we said earlier, this is possible, and such events are called independent events:
 If $P(E/F) = P(E)$ $\Rightarrow \,\,\,\,E$ and $F$ are independent events
Let us see an example. In the card-drawing experiment, let events $E$ and $F$ be defined as
 $E$ : The card has a number greater than $8$ $F$ : The card is black
The alert reader will immediately realize that $P(E/F)$ is the same as $P(E)$. Why? Because, the knowledge that the card is black does not change the number of cards per suit that are greater than eight. Stated explicitly,
 $P(E) = \dfrac{{\{ 9,\;10,\;J,\;Q,\;K\} \;{\rm{per}}\;{\rm{suit}}\; \times \;4\;{\rm{suits}}}}{{52}}$ $= \dfrac{{20}}{{52}} = \dfrac{5}{{13}}$ and $P(E/F) = \dfrac{{\{ 9,\;10,\;J,\;Q,\;K\} \;{\rm{per}}\;{\rm{suit}}\; \times 2\;{\rm{suits}}}}{{26}}$ because there are only two black suits $= \dfrac{5}{{13}}$
There is one important point the reader must notice and appreciate:
If $P(E/F) = P(E)$
 $\rightarrow$ this means that $E$ and $F$ are independent events $\rightarrow$ this should also mean that $P(E/F)$ should be the same as $P(F)$
Let us verify this in the example above. We have,
 $P(F) = \dfrac{{\# \;{\rm{black}}\;{\rm{cards}}}}{{\# \;{\rm{total}}\;{\rm{cards}}}}$ $= \dfrac{{26}}{{52}} = \dfrac{1}{2}$
and,
 $P(F/E) = \dfrac{{\# \;{\rm{black}}\;{\rm{cards}}\;{\rm{greater}}\;{\rm{than}}\;{\rm{eight}}}}{{\# \;{\rm{total}}\;{\rm{cards}}\;{\rm{greater}}\;{\rm{than}}\;{\rm{eight}}}}$ $= \dfrac{{\{ 9,\;10,\;J,\;Q,\;K\} \;{\rm{of}}\;{\rm{Spades}}\;{\rm{and}}\;{\rm{of}}\;{\rm{Clubs}}}}{{\{ 9,\;10,\;J,\;Q,\;K\} \;{\rm{per}}\;{\rm{suit}}\; \times \;4\;{\rm{suits}}}}$ $= \dfrac{{10}}{{20}}$ $= \dfrac{1}{2}$
which confirms the assertion
 $\begin{array}{l} \;{\rm{IF}}\;A\;{\rm{and}}\;B\;{\rm{are}}\;{\rm{independent}}\;{\rm{events}}\;\\ \;\;\;\;\;\;\;\;\; \Rightarrow \,\,\,P(A/B) = P(A)\\ \;\;\;\;\;\;\;\;\,{\rm{and}}\;P(B/A) = P(B) \end{array}$ $\ldots(1)$
Note that the favorable cases while calculating $P(A/B)$ are those cases in $A$that are common to $B$; the total cases are all cases in $B$. Thus
 $P(A/B) = \dfrac{{\# \;{\rm{favorable}}\;{\rm{cases}}}}{{\# \;{\rm{total}}\;{\rm{cases}}}}$ $= \dfrac{{\# \;(A \cap B)}}{{\# \;(B)}}$ $\ldots(2)$
If the entire sample space of the experiment consists of $N$ out comes, we can write $(2)$ as
 $P(A/B) = \dfrac{{\# \;\left( {A \cap B} \right)/N}}{{\# \;\left( B \right)/N}}$ $= \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$ $\ldots(3)$
Similarly,
 $P\left( {B/A} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$ $\ldots(4)$
This means that if $A$ and $B$ are independent, then
 $P\left( {A/B} \right) = P(A) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$ From $(1)$ and $(3)$ ${P\left( {A \cap B} \right) = P(A) \cdot P(B)}$
Thus, the probability of the intersection of two independent events is simply the product of the individual probabilities.