Probability
6.0 Multiplication Theorem
6.0 Multiplication Theorem
$$P(A \cap B) = P(B).P(A/B) = P(A).P(B/A)$$
Proof: Let $S$ be a sample space, and have total of $n$ outcomes possible.
Let $p$, $q$ and $r$ be the number of outcomes favourable to the events $A$, $B$, and $A \cap B$
Then,
$$\begin{equation} \begin{aligned} P(A) = {{n(A)} \over {n(S)}} = {p \over n} \\ P(B) = {{n(B)} \over {n(S)}} = {q \over n} \\ P(B/A) = {{n(A \cap B)} \over {n(A)}} = {r \over p} \\ P(A/B) = {{n(A \cap B)} \over {n(B)}} = {r \over q} \\\end{aligned} \end{equation} $$
Now, consider
$$P(A \cap B) = {{n(A \cap B)} \over {n(S)}} = {r \over n}$$
Multiplying and dividing by $p$,
$$\begin{equation} \begin{aligned} P(A \cap B) = {r \over p} \times {p \over n} \\ P(A \cap B) = P(B/A) \times P(A) \\\end{aligned} \end{equation} $$
Similarly,
Multiplying and dividing by $q$,
$$\begin{equation} \begin{aligned} P(A \cap B) = {r \over q} \times {q \over n} \\ P(A \cap B) = P(A/B) \times P(B) \\\end{aligned} \end{equation} $$
Extension of the multiplication theorem
$$P({A_1} \cap {A_2} \cap {A_3}..... \cap {A_n}) = P(\,{A_1})P(\,{A_2}/{A_1})P(\,{A_3}/{A_1} \cap {A_2})......P(\,{A_n}/{A_1} \cap {A_2} \cap {A_3}..... \cap {A_{n - 1}})$$
Illustration 26. A bag contains $10$ white and $15$ black balls. Two balls are drawn in succession without replacement. What is the probability that the first is white and the second is black?
Solution: Let us consider the events to be,
$A$ - Getting a white ball in the first draw
$B$ - Getting a black in second draw
$$P(A) = \;{{^{10}{C_1}} \over {^{25}{C_1}}} = {{10} \over {25}} = {2 \over 5}$$
Since the ball is not replaced, the event $B$ occurs on a condition that $A$ has already occurred.
When $A$ has already occurred, the number of white balls left will be $9$, black balls will be $15$ and hence there will be $24$ balls.
$$P(B/A) = \;{{^{15}{C_1}} \over {^{24}{C_1}}} = {{15} \over {24}} = {5 \over 8}$$
Probability of getting white ball in the first draw and black is the second draw = probability of both the events occurring.
= $P(A \cap B)$
$$\begin{equation} \begin{aligned} P(A \cap B) = P(B/A) \times P(A) \\ P(A \cap B) = {5 \over 8} \times {2 \over 5} = {1 \over 4} \\\end{aligned} \end{equation} $$
Thus the probability of getting a white ball in the first draw and black in the second draw is ${1 \over 4}$.
Illustration 27. Two balls are drawn from an urn containing $2$ white, $3$ red and $4$ black balls one by one without replacement. What is the probability that at least one ball is red?
Solution: Let us consider the following events,
$F$ = Getting red ball in the first draw.
$S$ = Getting red ball in the second draw.
The probability of getting red balls in at least one draw = 1- not getting red balls in both the draws.
i.e.
$$P(F \cup S) = 1 - P\overline {(F \cup S)} $$
$$P(F \cup S) = 1 - P(\bar F \cap \bar S)$$ [By De Morgan's law]
$P(\bar F)$ = Probability of not getting red in the first draw.
i.e. Probability of getting any other color in the first draw.
$ = {6 \over 9} = {2 \over 3}$
Event ${\bar S}$ happens on a condition that event ${\bar F}$ has already occurred.
Thus, $P(\bar S/\bar F) = $ not getting red in the second draw given, not getting red in the first draw.
After ${\bar F}$ occurs, there are $5$ balls of other colors and in total $8$ balls.
$$P(\bar S/\bar F) = {5 \over 8}$$
$$P(F \cup S) = 1 - P(\bar F \cap \bar S) = 1 - P(\bar F)P(\bar S/\bar F)$$
$$P(F \cup S) = 1 - {2 \over 3} \times {5 \over 8} = 1 - {5 \over {12}} = {7 \over {12}}$$
The probability of getting red ball in at least one of the draws is ${7 \over {12}}$
Question 18. A bag contains $n$ white and $n$ black balls. Pairs of balls are drawn without replacement until the bag is empty. Find the probability that each pair consists of one white and one black ball.
Solution: Let ${A_i}\;(i = 1,2,...,n)$ denote the event of getting one white and one black ball in the ${i^{th}}$ draw.
Then, getting one white and one black in every draw of a pair is denoted as ${A_1} \cap {A_2} \cap {A_3} \cap ..... \cap {A_n}$
$$P({A_1} \cap {A_2} \cap {A_3} \cap ..... \cap {A_n}) = P({A_1})P({A_2}/{A_1})P({A_3}/{A_1} \cap {A_2}).....P({A_n}/{A_1} \cap {A_2} \cap ... \cap {A_{n - 1}})$$
$P({A_1})$ = Getting one white and one black on the first draw of pair of balls
$$P({A_1}) = {{^n{C_1}{ \times ^n}{C_1}} \over {^{2n}{C_2}}} = {{{n^2}} \over {^{2n}{C_2}}}$$
$P({A_2}/{A_1})$ = Getting a one white and one black in the second draw, given that the same kind occurred in the first draw
After the first draw, the number of white balls and black balls would be $n - 1$ each, and the total number of balls will be $2n - 2$.
$$P({A_2}/{A_1}) = {{^{n - 1}{C_1}{ \times ^{n - 1}}{C_1}} \over {^{2n - 2}{C_2}}} = {{{{(n - 1)}^2}} \over {^{2n - 2}{C_2}}}$$
$$P({A_3}/{A_1} \cap {A_2}) = {{^{n - 2}{C_1}{ \times ^{n - 2}}{C_1}} \over {^{2n - 4}{C_2}}} = {{{{(n - 2)}^2}} \over {^{2n - 4}{C_2}}}$$
and so on.
Finally,
$$P({A_{n - 1}}/{A_1} \cap {A_2} \cap ... \cap {A_{n - 2}}) = {{^2{C_1}{ \times ^2}{C_1}} \over {^4{C_2}}} = {{{2^2}} \over {^4{C_2}}}$$
$$P({A_n}/{A_1} \cap {A_2} \cap ... \cap {A_{n - 1}}) = {1 \over {^2{C_2}}}$$
$$\begin{equation} \begin{aligned} P({A_1})P({A_2}/{A_1})P({A_3}/{A_1} \cap {A_2}).....P({A_n}/{A_1} \cap {A_2} \cap ... \cap {A_{n - 1}}) = {{{n^2}} \over {^{2n}{C_2}}} \times {{{{(n - 1)}^2}} \over {^{2n - 2}{C_2}}} \times {{{{(n - 2)}^2}} \over {^{2n - 4}{C_2}}} \times .... \times {{{2^2}} \over {^4{C_2}}} \times {1 \over {^2{C_2}}} \\ = {{{{(n!)}^2}} \over {{{(2n)(2n - 1)} \over 2} \times {{(2n - 2)(2n - 3)} \over 2} \times ..... \times {{4 \times 3} \over 2} \times {{2 \times 1} \over 2}}} \\ = {{{{(n!)}^2}{2^n}} \over {(2n)!}} \\ = {{{2^n}} \over {^{2n}{C_n}}} \\\end{aligned} \end{equation} $$
Hence, the probability is $${{{2^n}} \over {^{2n}{C_n}}}$$
