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}}.

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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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