Understanding Multiplication Rule & Independent/Dependent Events in Elementary Statistics, Study notes of Statistics

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Multiplication Rule

in

Probability

with

Tree Diagram

What is the Multiplication Rule?

Given two different events A and B, the Multiplication Rule states

P(A and B) = P(A) · P(B|A) Where ◮ (^) P(A and B) implies that event A happens first, then event B is to happen, and ◮ (^) and P(B|A) implies that event A has taken place already, therefore P(B) has to be adjusted accordingly if necessary.

Solution:

An ordinary full-deck of playing cards has 52 cards and 12 of them are face cards as shown here.

Let F 1 be the event that the first card is a face card, and F 2 be the event that the second card is a face card, using the multiplication rule, we need to evaluate the following formula.

P(F 1 and F 2 ) = P(F 1 ) · P(F 2 |F 1 )

Solution Continued:

Working with a full-deck of playing cards, P(F 1 ) =

, however

when calculating P(F 2 |F 1 ), we need to know what happens to the first card. ◮ (^) With replacement ⇒ P(F 2 |F 1 ) =

◮ (^) Without replacement ⇒ P(F 2 |F 1 ) =^11 51 so ◮ (^) With replacement ⇒ P(F 1 and F 2 ) = P(F 1 ) · P(F 2 |F 1 ) =

◮ (^) Without replacement ⇒ P(F 1 and F 2 ) = P(F 1 ) · P(F 2 |F 1 ) =

Example:

Given: P(A) = 0. 6 , P(B) = 0.5, and A and B are independent events, find P(A and B) and P(A or B).

Solution:

Since A and B are independent events, we get

P(A and B) = P(A) · P(B) = 0. 6 · 0 .5 = 0. 3

Using the addition rule, we get

P(A or B) = P(A) + P(B) − P(A and B) = 0 .6 + 0. 5 − 0 .3 = 0. 8

What are Dependent Events?

Two different events A and B, are called Dependent Events means that the probability of one event is affected by the result of an earlier event.

What is the property of Dependent Events?

For any two dependent events A and B,

P(B|A) = P(A and B) P(A)

Example:

The probability that a freshman taking a math class is 0.75. The probability of taking a math class and an English class is 0.4. What is the probability that a randomly selected freshman taking an English class given that he or she is taking a math class?

Solution:

Let M be the event that a randomly selected freshman is taking a math class and E be the event that a randomly selected freshman is taking an English class. So P(M) = 0.75 and P(M and E ) = 0.4, now we can use the conditional probability formula to find P(E |M).

P(E |M) =

P(M and E ) P(M) =

What is a Tree Diagram?

Tree diagrams are a helpful tool for calculating probabilities when there are several events involved.

How do we use the Tree Diagram?

◮ (^) The probability of each branch is written on the branch. ◮ (^) The outcome is written at the end of the branch. ◮ (^) Multiply probabilities along the branches. ◮ (^) Add probabilities of those branches that satisfy the desired event.

Example:

Use tree diagram to display all outcomes with indicated probabilities for each branch when you toss three fair coins.

Solution:

Example:

Suppose a fair coin is tossed three times,

◮ (^) find the probability of getting all tails. ◮ (^) find the probability of getting all heads. ◮ (^) find the probability of getting all tails or all heads. ◮ (^) find the probability of getting neither all tails nor all heads.

Solution:

Using the tree diagram, we get

P( All Tails ) =

Example:

An ordinary full-deck of playing cards has 52 cards and 12 of them are face cards as shown below.

Use the tree diagram to find the probability of all outcomes of selecting two face cards without replacement.

Solution:

Let F 1 be the event that the first card is a face card, and F 2 be the event that the second card is a face card, and similarly Let F¯ 1 be the event that the first card is not a face card, and F¯ 2 be the event that the second card is not a face card.

Drawing Two Face Cards (No Replacement)

F

F¯

F

F F¯

F^ ¯

F F¯

F 1 F 2 F 1 F¯ 2 F¯ 1 F 2 F¯ 1 F¯ 2

12 52 11 51

40 51

40 52 12 51

39 51