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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 and B)=P(A)×P(B)~~~~~ P(A \text{ and } B) = P(A) \times P(B)

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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 and ace)=P(queen)×P(ace)=452×452=162704=1169P(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 SS denote stress. ThenP(S and S and S)=(0.46)3=0.097.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 TT denote a positive test result. ThenP(T and T and T and T)=(0.32)4=0.010.P(T \text{ and } T \text{ and } T \text{ and } T) = (0.32)^4 = 0.010.

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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 BB given AA is P(B∣A)P(B|A).
This does not mean B÷AB \div A

When two events A and B are dependent the probability of both occurring is
     P(A and B)=P(A)×P(B∣A)~~~~~ P(A \text{ and } B) = P(A) \times P(B|A)

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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(D1 and D2)=P(D1)×P(D2∣D1)=225×124=2600=1300P(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 and A)=P(H)×P(A∣H)=0.53×0.27=0.1431P(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. Box 1 and Box 2
Solution:P(red)=P(red from box 1)+P(red from box 2)=(12)(23)+(12)(14)=26+18=824+324=1124P(\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}

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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)=P(A and B)P(A)~~~~~ P(B|A) = \dfrac{P(A \text{ and } B)}{P(A)}

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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 NN = parking in a no parking zone and
      T~~~~~~T = getting a ticket.
Then,
     P(T∣N)=[P(T and N)]P(N)=0.060.2=0.30~~~~~ 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.
GenderYesNoTotal
Male321850
Female84250
Total4060100
Find the probability that the respondent answered yes given that the respondent was a female.
Solution:Let MM = respondent was a male
      F~~~~~~F = respondent was a female
      Y~~~~~~Y = respondent answered “yes”
      N~~~~~~N = respondent answered “no”P(Y∣F)=P(F and Y)P(F)=[8100][50100]=425P(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)=P(N and M)P(N)=[18100][60100]=310P(M|N) = \dfrac{P(N \text{ and } M)}{P(N)} = \dfrac{[\dfrac{18}{100}]}{[\dfrac{60}{100}]} = \dfrac{3}{10}