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Independent Events. Slideshow 54, Mathematics Mr Richard Sasaki, Room 307. Recalling the meaning of “with replacement” and “without replacement” Understand independence and calculating probabilities about 2 events. Objectives. We need a bit of a review. Vocabulary.

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Independent events

Independent Events

Slideshow 54, Mathematics

Mr Richard Sasaki, Room 307


Recalling the meaning of “with replacement” and “without replacement”

Understand independence and calculating probabilities about 2 events

Objectives


We need a bit of a review

We need a bit of a review. “without replacement”

Vocabulary

The thing that is taking place (eg: Rolling a die)

Event (Trial) -

Possible outcomes for the event (for a die: 1, 2, 3, 4, 5, 6)

Value -

Frequency -

The number of times a value appears in an experiment.


In the winter homework independence was mentioned what is it again

In the Winter Homework, “without replacement”Independence was mentioned. What is it again?

Independence

Independence for events is where one event doesn’t affect another.

This means that no matter what happens in one event, the probabilities for the other event are exactly the same.

Last lesson we looked at pulling objects out of a bag and “with replacement” and “without replacement”. Let’s review those meanings.


With replacement

With Replacement – “without replacement”

With and Without Replacement

After an event occurs, everything is “reset” (put back as it was) so when we repeat, nothing has changed.

Without Replacement –

After an event occurs, whatever happened is removed from the event, causing all future occurrences to have differing probabilities.

Which of these shows independence?

With Replacement


Let s consider flipping an unbiased coin twice

Let’s consider flipping an unbiased coin twice. “without replacement”

Flipping a Coin…Twice!

What are the possibilities we can get?

You will be expected to list them…

Heads, Heads

Note – Heads, Tails is different to Tails, Heads. We get one of each but order matters here.

Heads, Tails

Tails, Heads

Tails, Tails


Flipping a coin twice
Flipping a Coin…Twice! “without replacement”

Heads, Heads

P(H, H) =

Heads, Tails

P(H, T) =

Tails, Heads

P(T, H) =

Tails, Tails

P(T, T) =

What are the probabilities of each of these occurring?

Why does each outcome have a probability of ¼?

# of successes

=

Total # of outcomes


Flipping a coin twice1
Flipping a Coin…Twice! “without replacement”

Heads, Heads

P(H, H) =

Heads, Tails

P(H, T) =

Tails, Heads

P(T, H) =

Tails, Tails

P(T, T) =

How about the following?

Order isn’t mentioned.

P(H and T) = P(H, T) + P(T, H).

P(A Heads and a Tails) =

We always get heads or tails!

P(A Heads or a Tails) =


Answers easy
Answers - Easy “without replacement”

No. Both events are independent with the same probabilities for each outcome.

P(Tails) = ½

Yes, their outcomes don’t affect each other.

1, 2, 3

P(2) =

1, 1 2, 1 3, 1

1, 2 2, 2 3, 2

1, 3 2, 3 3, 3

P(H, T) = ¼

P(At least one heads) =

P(1, 3) =

P(Exactly one heads) =

P(2, 2) =

P(No tails) =

P(No 4) =


Answers hard
Answers - Hard “without replacement”

One more question…

P(Both Even) =

16

P(Both less than or equal to 4) =

P(2, 4) =

12

P(A 1 and a 3) =

P(H, 4) =

P(Exactly one 2) =

P(Tails and greater than 2) =

P(Total of 7) =

P(Two sixes) = 0

P(Total of less than 3) =

Because order is considered. The coin is flipped first so we can’t flip a 3.

P(A 3 and a 4) =

P(Total of 7) =


An introduction to permutations
An Introduction to Permutations “without replacement”

Many of our examples involved choosing 2 from a group of some number of options with repetition allowed (picking the same number twice).

Two options, two picked (Coin) -

4

Three options, two picked (Spinner) -

9

Four options, two picked -

16

Six options, two picked -

36

So for options, if we pick two with repetition, we get combinations.


An introduction to permutations1
An Introduction to Permutations “without replacement”

Permutations are combinations where order matters. (These are like the ones we did today.)

To calculate permutations with repetition where order matters for possible values and choosing of them, we get…

So if we rolled a 10 sided die four times, how many combinations exist?


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