Different to the case of  __ independent probability__.

Examples which involve dependent probability, look at when the probability of an event outcome

*is*influenced by the outcome of another event.

Which are times when we can make use of a probability of dependent events formula.

If we flip a coin, each flip/event that is always independent of the previous flip.

The outcome of a new coin flip is NOT affected by the outcome of any previous flips that happened before.

We can look at an example where event outcomes are NOT independent from one another.

__Example    __

*(1.1) *Say we were to pick out  2  cards from a normal deck of  52  playing cards.

We want to know the probability of pulling out a Spade card on the second pick.

Now there are  52  cards in a standard playing card deck.

From these cards,  13  of the cards are a Spade,

so  39  of the cards are

*not*a Spade card.

Probability of a Spade in first pick:    

**P**(

**Spade 1st**)  =  \bf{\frac{13}{52}}

### How is Probability of Spade on 2nd pick affected?

After the first card has been picked, the probability of picking out a Spade card with the second pick will be affected by the outcome of the first pick.

**–**  If the first card picked is a Spade,

there is  1  less Spade card, and  51  cards remaining in the pack overall.

Then the probability of a Spade card on  2nd  pick is:

**P**(

**Spade 2nd**)  =  \bf{\frac{12}{51}}

**–**  If the first card is NOT a Spade card,

there is still  13  Spade cards, and  51  cards in the whole pack.

Then the probability of a Spade card on  2nd  pick is:

**P**(

**Spade 2nd**)  =  \bf{\frac{13}{51}}

As can be observed with this example, the probability of picking out a Spade card on the  2nd  pick is dependent on the previous outcome of the  1st  card drawn.

Situations like this involve dependent probability.

## Dependent Probability Notation

and Examples

### Dependent Probability Notation,

Probability of Dependent Events Formula

If we have an event, let’s call this event  

**G**.

**P**(

**G**)  =  Probability of  

**G**

Now say there is a  2nd  event, we can call this event  

**H**.

**P**(

**H**)  =  Probability of  

**H**

Here we introduce some added notation for dependent probability which is:

**P**(

**G | H**)  =  Probability of event  H,  given event  G  has already happened.

Notation which means that event  G  definitely has occurred, now what is the probability of event  H  occurring.

Thus if  2  events  G  and  H  are to happen, and probability of event  H is dependent on the outcome of event  S,  the sum for dependent probability is:

**P**(

**G**∩

**H**)  =  

**P**(

**G**) ×

**P**(

**G | H**)

__Example    __

*(2.1) ***10**Red balls   ,  

**10**Black balls   ,  

**20**White balls

Someone is going to pick  3  balls at random out of the rucksack.

What is the probability that they pick out  3  balls that are all Black?

*Solution*   Thinking in terms of a Black ball being picked each time, we have.

Probability of Black on first pick:  

**P**(

**1st Black**)  =  \bf{\frac{10}{40}}

Probability of Black on second pick:  

**P**(

**2nd Black**)  =  \bf{\frac{9}{39}}

Probability of Black on second pick:  

**P**(

**3rd Black**)  =  \bf{\frac{8}{38}}

Now:

Probability of a Black ball on first and second pick:

**P**(

**2 Black**)  = 

**P**(

**1st Black**) ×

**P**(

**2nd Black**|

**1st Black**)  =  \bf{\frac{10}{40}} × \bf{\frac{9}{39}}   =   \bf{\frac{90}{1560}}   =   \bf{\frac{3}{52}}

Probability of Black on first, second and third pick:

**P**(

**3 Blacks**)  = 

**P**(

**2 Blacks**) ×

**P**(

**3rd Black**|

**2 Blacks**)  =  \bf{\frac{3}{52}} × \bf{\frac{8}{38}}   =   \bf{\frac{24}{1976}}   =   \bf{\frac{3}{247}}

The probability of someone picking out  3  Black balls in a row from the original  40  billiard balls in the rucksack is  \bf{\frac{3}{247}},  which amounts to roughly  0.012.

Not very likely to happen.

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