Multiplicative Rules

Multiplying the formula of Definition 2.9 by , we obtain the multiplicative rule, which enables us to calculate the probability that two events will both occur.
Theorem 2.13:
$\fbox{\parbox{5in}{ If in an experiment the events $A$\ and $B$\ can both occur,... ...playmath} P(A \cap B) = P(A)*P(B\vert A) \end{displaymath}provided $P(A) > 0$ }}$
We can also write

$\displaystyle P(A \cap B) = P(B \cap A) = P(B)*P(A\vert B)$
Example 2.35: Suppose that we have a fuse box containing 20 fuses, of which 5 are defective. If 2 fuses are selected at random and removed from the box in succession without replacing the first.
What is the probability that both fuses are defective?
- Event : the first fuse is defective
- Event : the second fuse is defective. Hence,
  
  $\displaystyle P(A \cap B) = P(A)*P(B\vert A) = \frac{1}{4}*\frac{4}{19} = \frac{1}{19}$

Example 2.36: One bag contains 4 white balls and 3 black balls. A second bag contains 3 white balls and 5 black balls.
One ball is drawn from the first bag and placed unseen in the second bag. What is the probability that a ball now drawn from the second bag is black?
Solution: Let , , and represent, respectively, the drawing of a black ball from bag 1, a black ball from bag 2, and a white ball from bag 1.

$\displaystyle p[(B_1 \cap B_2) \cup ( W_1 \cap B_2)]=P(B_1 \cap B_2) + P(W_1 \cap B_2)$

$\displaystyle =P(B_1)P(B_2\vert B_1)+P(W_1)P(B_2\vert W_1)$

$\displaystyle =\frac{3}{7}*\frac{6}{9}+\frac{4}{7}*\frac{5}{9}=\frac{38}{63}$

**Figure 2:** Tree diagram for Example 2.36.
$\includegraphics[scale=0.35]{figures/02-06}$

Theorem 2.14:
$\fbox{\parbox{5in}{ Two events $A$\ and $B$\ are (statistically or probabilistic... ...ill both occur, we simply find the product of their individual probabilities. }}$
Example 2.37: A small town has one fire engine and one ambulance available for emergencies.
- The probability that the fire engine is available when needed is 0.98,
- The probability that the ambulance is available when called is 0.92
- In the event of an injury resulting from a burning building, find the probability that both the ambulance and the fire engine will be available.
Solution: Let and represent the respective evens that the fire engine and the ambulance are available. Then

$\displaystyle P(A \cap B) = P(A)P(B) = 0.98 * 0.92 = 0.9016.$

Example 2.38: Find the probability that

Figure 3: An electrical system for Example 2.38.

$\includegraphics[scale=0.5]{figures/02-07}$
- the entire system works
- the component does not work, given that the entire system works
Solution:

$\displaystyle P(A \cap B \cap (C \cup D))= P(A)*P(B)*P(C \cup D)$

$\displaystyle =P(A)*P(B)*(1-P(C' \cap D'))=P(A)*P(B)*(1-P(C')*P(D'))$

$\displaystyle =0.9*0.9*(1-(1-0.8)*(1-0.8))=0.7776$
$\displaystyle P=\frac{P(the~system~works~but~C~does~not~work)}{P(the~system~works)}$

$\displaystyle =\frac{P(A \cap B \cap C' \cap D)}{P(A \cap B \cap (C \cup D))}=\frac{0.9*0.9*(1-0.8)*0.8}{0.7776}=0.1667$

**Figure 3:** An electrical system for Example 2.38.
$\includegraphics[scale=0.5]{figures/02-07}$

Independence is often easy to grasp intuitively.
For example, if the occurrence of two events is governed by distinct and non-interacting physical processes, such events will turn out to be independent.
On the other hand, independence is not easily visualized in terms of the sample space.
A common fallacy (wrong idea) is that two events are independent if they are disjoint, but in fact the opposite is true:
$\fbox{\parbox{5in}{ Two disjoint events $A$\ and $B$\ with $P(A) > 0$\ and $P(B)... ...pendent, since their intersection $A \cap B$\ is empty and has probability 0. }}$
We note that

(i)

independent events are never mutually exclusive,

(ii)

two mutually exclusive events are always dependent.

Theorem 2.15:
$\fbox{\parbox{5in}{ If the events $A_1,A_2,A_3,\ldots ,A_k$\ can occur, then \be... ...ots \cap A_k)=P(A_1)P(A_2)\ldots P(A_k)=\prod_{n=1}^k P(A_n) \end{displaymath}}}$
Example 2.39: Three cards are drawn in succession without replacement. Find the probability that the event $A_1 \cap A_2 \cap A_3$ occurs, where
- : the first card is red ace
- : the second card is a 10 or jack
- : the third card is greater than 3 but less than 7
Solution:

$\displaystyle P(A_1 \cap A_2 \cap A_3)=P(A_1) P(A_2\vert A_1) P(A_3 \vert A_1 \cap A_2)$

$\displaystyle =\frac{2}{52}*\frac{8}{51}*\frac{12}{50}=\frac{8}{5525}$

Independence of Several Events:
$\fbox{\parbox{5in}{ The events $A_1,A_2,A_3,\ldots ,A_n$\ are \textbf{independen... ...{i\in S} P(A_i) \end{displaymath}for any subset $S$\ of $\{1,2,\ldots ,n\}$. }}$
Independence means that the occurrence or non-occurrence of any number of the events from that collection carries no information on the remaining events or their complements.
Example: Independence of three events: If and are independent,

$\displaystyle P(A_1 \cap A_2)=P(A_1)P(A_2)$

$\displaystyle P(A_1 \cap A_3)=P(A_1)P(A_3)$

$\displaystyle P(A_2 \cap A_3)=P(A_2)P(A_3)$

$\displaystyle P(A_1 \cap A_2 \cap A_3)=P(A_1)P(A_2)P(A_3)$

Example: Consider two independent fair coin tosses, and the following events:
- = $1^{st}$ toss is a head,
- = $2^{nd}$ toss is a head,
- = the two tosses have different results.
Pairwise independence does not imply independence.
- and are independent, by definition.
- $P(D\vert H_1)= P(D)$ and $P(D\vert H_2)= P(D)$
- $P(H_1 \cap H_2 \cap D)=0 \neq P(H_1)P(H_2)P(D)$
Example: Consider two independent rolls of a fair die, and the following events:
- = $1^{st}$ roll is 1, 2, or 3, = $2^{nd}$ roll is 3, 4, or 5, = the sum of the two rolls is 9.
$P(A_1 \cap A_2 \cap A_3) =P(A_1)P(A_2)P(A_3)$ is not enough for independence.
- $P(A \cap B) = \frac{1}{6} \neq \frac{1}{2} *\frac{1}{2} = P(A)P(B)$
- $P(A \cap C) = \frac{1}{36} \neq \frac{1}{2} *\frac{4}{36} = P(A)P(C)$
- $P(B \cap C) = \frac{3}{6} \neq \frac{1}{2} *\frac{4}{36} = P(B)P(C)$
- $P(A \cap B \cap C) = \frac{1}{36} \neq \frac{1}{2} *\frac{1}{2}*\frac{4}{36} = P(A)P(B)P(C)$

Cem Ozdogan 2010-03-06