# Probability of Compound Events Examples

When we try to solve the probability of compound events examples, we are aiming to work out the probability of  2  events occurring, instead of just  1  event.

These  2  events can be either  “mutually exclusive”,  or  “mutually inclusive”.

### Mutually Exclusive Events

When events as classed as mutually exclusive, they are events that can’t occur at the same time.

For example if you were flipping a coin you can get either a &nbsp”Head”&nbsp or a &nbsp”Tails”,&nbsp you can’t get both a &nbspHead&nbsp and a &nbspTail&nbsp on one flip of a coin.

A &nbspHead&nbsp and a &nbspTail&nbsp occurring on a coin flip are mutually exclusive events.

When you have &nbsp2&nbsp mutually exclusive events &nbspA&nbsp and &nbspB,&nbsp the probability of either occurring can be worked out in the following way.

P( A or B ) &nbsp = &nbsp P( A ) + P( B)

Examples &nbsp &nbsp

(1.1)&nbsp

On rolling of a fair dice, what is the probability of rolling a &nbsp1&nbsp or a &nbsp4?

Solution&nbsp &nbsp

P( 1 ) &nbsp=&nbsp \bf{\frac{1}{6}} &nbsp , &nbsp P( 4 ) &nbsp=&nbsp \bf{\frac{1}{6}}

P( 1 or 4 ) &nbsp=&nbsp \bf{\frac{1}{6}} + \bf{\frac{1}{6}} &nbsp = &nbsp \bf{\frac{2}{6}} &nbsp = &nbsp \bf{\frac{1}{3}}

The probability of landing either a &nbsp1&nbsp or a &nbsp4&nbsp on one dice roll is &nbsp\bf{\frac{1}{3}}.

(1.2)&nbsp

A bag of golf balls has &nbsp14&nbsp balls inside numbered from &nbsp1&nbsp to &nbsp14.

What is the probability of randomly selecting a golf ball from the bag that has a number less than &nbsp5&nbsp or higher than &nbsp11?

Solution&nbsp &nbsp

P( less than 5 ) &nbsp=&nbsp \bf{\frac{4}{14}}

P( higher than 11 ) &nbsp=&nbsp \bf{\frac{3}{14}}

P( less than 5 &nbspor&nbsp higher than 11) &nbsp=&nbsp \bf{\frac{4}{14}} + \bf{\frac{3}{14}} &nbsp = &nbsp \bf{\frac{7}{14}} &nbsp = &nbsp \bf{\frac{1}{2}}

The probability of picking out a golf ball which is numbered either less than &nbsp5&nbsp or higher than &nbsp11&nbsp is &nbsp\bf{\frac{1}{2}}.

## Probability of Compound Events Examples,Mutually Inclusive Events

With probability of compound events examples,&nbsp “mutually inclusive events”&nbsp are events that can occur at the same time as each other.

If we recall the bag of numbered golf balls from example &nbsp(1.2)&nbsp again.

We can consider the probability of picking out a ball numbered less than &nbsp8,&nbsp or a ball specifically numbered &nbsp4.

These events could both occur at the same time, as a golf ball numbered &nbsp4,&nbsp is also a golf ball that is numbered less than &nbsp8.
They are mutually inclusive events.

Establishing the probability of compound events cases such as this, involves the formula from before seen above, but with a slight alteration.

P( A or B ) &nbsp=&nbsp P( A ) + P( B ) &nbsp−&nbsp P( A and B )

### Golf ball either numbered 4 or less than 8:

For the probability of picking out a golf ball that is either a number &nbsp4&nbsp or a number less than an &nbsp8&nbsp.

P( 4 ) &nbsp=&nbsp \bf{\frac{1}{14}} &nbsp , &nbsp P( less than 8 ) &nbsp=&nbsp \bf{\frac{7}{14}}

P( 4 &nbsp AND &nbsp less than 8 ) &nbsp = &nbsp P( 4 ) × P( 4 &nbsp|&nbsp less than 8 ) &nbsp = &nbsp \bf{\frac{1}{14}} × 1 &nbsp = &nbsp \bf{\frac{1}{14}}

Now to work out the probability of picking out a golf ball that is either a &nbsp4&nbsp or less than an &nbsp8?

P( 4 &nbspor&nbsp less than 8 ) &nbsp = &nbsp P( 4 ) + P( less than 8 ) &nbsp−&nbsp P( 4 &nbspAND&nbsp less than 8 )

=> &nbsp&nbsp P( 4 or less than 7 ) &nbsp = &nbsp \bf{\frac{1}{14}} + \bf{\frac{7}{14}}\bf{\frac{1}{14}} &nbsp = &nbsp \bf{\frac{7}{14}} &nbsp = &nbsp \bf{\frac{1}{2}}

Example &nbsp &nbsp

(2.1)&nbsp

With a deck of &nbsp52&nbsp playing cards, what is the probability of picking out at random either a &nbspSPADE&nbsp or an &nbspACE&nbsp card?

Solution&nbsp &nbsp

P( SPADE &nbspor&nbsp ACE ) &nbsp = &nbsp P( SPADE ) + P( ACE ) &nbsp−&nbsp P( SPADE &nbspand&nbsp ACE )

P( SPADE ) &nbsp = &nbsp \bf{\frac{13}{52}} &nbsp &nbsp , &nbsp &nbsp P( ACE ) &nbsp = &nbsp \bf{\frac{4}{52}} &nbsp = &nbsp \bf{\frac{1}{13}}

P( SPADE &nbsp|&nbsp ACE ) &nbsp=&nbsp \bf{\frac{1}{4}}

P( SPADE &nbspand&nbsp ACE ) &nbsp = &nbsp P( ACE ) &nbsp×&nbsp P( SPADE &nbsp|&nbsp ACE ) &nbsp = &nbsp \bf{\frac{1}{13}} × \bf{\frac{1}{4}} &nbsp = &nbsp \bf{\frac{1}{52}}

P( SPADE &nbspor&nbsp ACE ) &nbsp = &nbsp \bf{\frac{13}{52}} + \bf{\frac{4}{52}} &nbsp−&nbsp \bf{\frac{1}{52}} &nbsp = &nbsp \bf{\frac{16}{52}}

The probability of picking out at random a &nbspSPADE&nbsp card or an &nbspACE&nbsp card is &nbsp{\frac{16}{52}}.

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