# Probability of Dependent Events Formula

Different to the case of &nbspindependent 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 &nbsp &nbsp

(1.1)&nbsp

Say we were to pick out &nbsp2&nbsp cards from a normal deck of &nbsp52&nbsp playing cards.

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

Now there are &nbsp52&nbsp cards in a standard playing card deck.

From these cards, &nbsp13&nbsp of the cards are a Spade,
so &nbsp39&nbsp of the cards are not a Spade card.

Probability of a Spade in first pick: &nbsp &nbsp P( Spade 1st ) &nbsp=&nbsp \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.

&nbsp If the first card picked is a Spade,
there is &nbsp1&nbsp less Spade card, and &nbsp51&nbsp cards remaining in the pack overall.

Then the probability of a Spade card on &nbsp2nd&nbsp pick is:

P( Spade 2nd ) &nbsp=&nbsp \bf{\frac{12}{51}}

&nbsp If the first card is NOT a Spade card,
there is still &nbsp13&nbsp Spade cards, and &nbsp51&nbsp cards in the whole pack.

Then the probability of a Spade card on &nbsp2nd&nbsp pick is:

P( Spade 2nd ) &nbsp=&nbsp \bf{\frac{13}{51}}

As can be observed with this example, the probability of picking out a Spade card on the &nbsp2nd&nbsp pick is dependent on the previous outcome of the &nbsp1st&nbsp card drawn.

Situations like this involve dependent probability.

## Dependent Probability Notationand Examples

### Dependent Probability Notation,Probability of Dependent Events Formula

If we have an event, let’s call this event &nbspG.

P( G ) &nbsp=&nbsp Probability of &nbspG

Now say there is a &nbsp2nd&nbsp event, we can call this event &nbspH.

P( H ) &nbsp=&nbsp Probability of &nbspH

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

P( G | H ) &nbsp=&nbsp Probability of event &nbspH,&nbsp given event &nbspG&nbsp has already happened.

Notation which means that event &nbspG&nbsp definitely has occurred, now what is the probability of event &nbspH&nbsp occurring.

Thus if &nbsp2&nbsp events &nbspG&nbsp and &nbspH&nbsp are to happen, and probability of event &nbspH is dependent on the outcome of event &nbspS,&nbsp the sum for dependent probability is:

P( GH ) &nbsp= &nbsp P( G ) × P( G | H )

Example &nbsp &nbsp

(2.1)&nbsp

A rucksack contains &nbsp40&nbsp equal sized billiard balls inside it, which are a follows.

10 Red balls &nbsp , &nbsp 10 Black balls &nbsp , &nbsp 20 White balls

Someone is going to pick &nbsp3&nbsp balls at random out of the rucksack.
What is the probability that they pick out &nbsp3&nbsp balls that are all Black?

Solution&nbsp &nbsp

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

Probability of Black on first pick: &nbsp P( 1st Black ) &nbsp=&nbsp \bf{\frac{10}{40}}

Probability of Black on second pick: &nbsp P( 2nd Black ) &nbsp=&nbsp \bf{\frac{9}{39}}

Probability of Black on second pick: &nbsp P( 3rd Black ) &nbsp=&nbsp \bf{\frac{8}{38}}

Now:

Probability of a Black ball on first and second pick:

P( 2 Black ) &nbsp=&nbsp P( 1st Black ) × P( 2nd Black | 1st Black ) &nbsp=&nbsp \bf{\frac{10}{40}} × \bf{\frac{9}{39}} &nbsp = &nbsp \bf{\frac{90}{1560}} &nbsp = &nbsp \bf{\frac{3}{52}}

Probability of Black on first, second and third pick:

P( 3 Blacks ) &nbsp=&nbsp P( 2 Blacks ) × P( 3rd Black | 2 Blacks ) &nbsp=&nbsp \bf{\frac{3}{52}} × \bf{\frac{8}{38}} &nbsp = &nbsp \bf{\frac{24}{1976}} &nbsp = &nbsp \bf{\frac{3}{247}}

The probability of someone picking out &nbsp3&nbsp Black balls in a row from the original &nbsp40&nbsp billiard balls in the rucksack is &nbsp\bf{\frac{3}{247}}, &nbspwhich amounts to roughly &nbsp0.012.

Not very likely to happen.

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