> ## Documentation Index
> Fetch the complete documentation index at: https://v1-learn.neoartd.my.id/llms.txt
> Use this file to discover all available pages before exploring further.
# The Multiplication Rules and Conditional Probability
Two events A and B are `independent` if the fact that A occurs does not affect the probability of B occurring.
> Dua peristiwa A dan B disebut `independen` jika fakta bahwa A terjadi tidak mempengaruhi probabilitas B terjadi.
**Example** : Rolling a die and getting a 6, and then rolling another die and getting a 3 are independent events.
When two events A and B
are independent the
probability of both
occurring is
$~~~~~ P(A \text{ and } B) = P(A) \times P(B) $
### Multiplication Rule 1 - Example
A card is drawn from a deck and
replaced; then a second card is
drawn. Find the probability of getting
a queen and then an ace.
Solution:
Because these two events
are independent (why?),
$P(queen \text{ and } ace) = P(queen) \times P(ace) = \dfrac{4}{52} \times \dfrac{4}{52} = \dfrac{16}{2704} = \dfrac{1}{169}$
A Harris pole found that 46% of
Americans say they suffer great stress
at least once a week. If three people
are selected at random, find the
probability that all three will say that
they suffer stress at least once a week.
Solution:
Let $S$ denote stress. Then
$P(S \text{ and } S \text{ and } S) = (0.46)^3 = 0.097.$
The probability that a specific medical
test will show positive is 0.32. If four
people are tested, find the probability
that all four will show positive.
Solution:
Let $T$ denote a positive test
result. Then
$P(T \text{ and } T \text{ and } T \text{ and } T) = (0.32)^4 = 0.010.$
## The Multiplication Rules and Conditional Probability
When the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way that the probability is changed, the events are said to be dependent.
> Ketika hasil atau kejadian dari peristiwa pertama mempengaruhi hasil atau kejadian dari peristiwa kedua sedemikian rupa sehingga probabilitas berubah, maka peristiwa tersebut dikatakan saling bergantung.
**Example** : Having high grades and getting a scholarship are dependent events.
The `conditional probability` of an event B in relationship to an event A is the probability that an event B occurs after event A has already occurred.
> Probabilitas kondisional dari suatu peristiwa B dalam hubungannya dengan peristiwa A adalah probabilitas bahwa peristiwa B terjadi setelah peristiwa A terjadi.
The notation for the conditional probability of $B$ given $A$ is $P(B|A)$.
This does not mean $B \div A$
When two events A and B
are dependent the
probability of both
occurring is
$~~~~~ P(A \text{ and } B) = P(A) \times P(B|A) $
### The Multiplication Rules and Conditional Probability - Example
In a shipment of 25 microwave ovens,
two are defective. If two ovens are
randomly selected and tested, find the
probability that both are defective if the
first one is not replaced after it has
been tested.
Solution:
Since the events are dependent,
$P(D_1 \text{ and } D_2) = P(D_1) \times P(D_2|D_1) = \dfrac{2}{25} \times \dfrac{1}{24} = \dfrac{2}{600} = \dfrac{1}{300}$
The WW Insurance Company found that
53% of the residents of a city had
homeowner's insurance with its company.
Of these clients, 27% also had automobile
insurance with the company. If a resident
is selected at random, find the probability
that the resident has both homeowner's
and automobile insurance.
Solution:
Since the events are dependent,
$P(H \text{ and } A) = P(H) \times P(A|H) = 0.53 \times 0.27 = 0.1431$
Box 1 contains two red balls and one
blue ball. Box 2 contains three blue
balls and one red ball. A coin is tossed.
If it falls heads up, box 1 is selected
and a ball is drawn. If it falls tails up,
box 2 is selected and a ball is drawn.
Find the probability of selecting a red
ball.
Solution:
$P(\text{red}) = P(\text{red from box 1}) + P(\text{red from box 2}) = (\dfrac{1}{2})(\dfrac{2}{3}) + (\dfrac{1}{2})(\dfrac{1}{4}) = \dfrac{2}{6} + \dfrac{1}{8} = \dfrac{8}{24} + \dfrac{3}{24} = \dfrac{11}{24}$
## Conditional Probability - Formula
The probability that the event B occurs
given that the first event A has occurred can be
found by dividing the probability that both events
occurred by the probability that the first event has
occurred The formula is
$~~~~~ P(B|A) = \dfrac{P(A \text{ and } B)}{P(A)} $
### Conditional Probability - Example
The probability that Sam parks in a no
parking zone `and` gets a parking ticket is
0.06, and the probability that Sam cannot
find a legal parking space and has to park
in the no-parking zone is 0.2. On Tuesday,
Sam arrives at school and has to park in a
no-parking zone. Find the probability that
he will get a ticket.
Solution:
Let $N$ = parking in a no parking zone and\
$~~~~~~T$ = getting a ticket.\
Then,\
$~~~~~ P(T|N) = \dfrac{[P(T \text{ and } N)]}{P(N)} = \dfrac{0.06}{0.2} = 0.30 $
A recent survey asked 100 people
if they thought women in the
armed forces should be permitted
to participate in combat.
| Gender | Yes | No | Total |
| ------ | :-: | :-: | :---: |
| Male | 32 | 18 | 50 |
| Female | 8 | 42 | 50 |
| Total | 40 | 60 | 100 |
Find the probability that the respondent
answered `yes` given that the respondent
was a female.
Solution:
Let $M$ = respondent was a male\
$~~~~~~F$ = respondent was a female\
$~~~~~~Y$ = respondent answered “yes”\
$~~~~~~N$ = respondent answered “no”
$P(Y|F) = \dfrac{P(F \text{ and } Y)}{P(F)} = \dfrac{[\dfrac{8}{100}]}{[\dfrac{50}{100}]} = \dfrac{4}{25}$
Find the probability that the respondent
was a male, given that the respondent
answered `no`.
Solution:
$P(M|N) = \dfrac{P(N \text{ and } M)}{P(N)} = \dfrac{[\dfrac{18}{100}]}{[\dfrac{60}{100}]} = \dfrac{3}{10}$