Question 207291
What is the probability of:
drawing a HEART
drawing an ACE
drawing the ACE of HEARTS 
Could you work these out so I have a greater understanding on these types of problems
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This is the sample space of 52 cards, 13 each of 
HEARTS, DIAMONDS, SPADES, and CLUBS:

A<font color="red">&hearts;</font>, 2<font color="red">&hearts;</font>, 3<font color="red">&hearts;</font>, 4<font color="red">&hearts;</font>, 5<font color="red">&hearts;</font>, 6<font color="red">&hearts;</font>, 7<font color="red">&hearts;</font>, 8<font color="red">&hearts;</font>, 9<font color="red">&hearts;</font>, 10<font color="red">&hearts;</font>, J<font color="red">&hearts;</font>, Q<font color="red">&hearts;</font>, K<font color="red">&hearts;</font>,

A<font color="red">&#9830</font>, 2<font color="red">&#9830</font>, 3<font color="red">&#9830</font>, 4<font color="red">&#9830</font>, 5<font color="red">&#9830</font>, 6<font color="red">&#9830</font>, 7<font color="red">&#9830</font>, 8<font color="red">&#9830</font>, 9<font color="red">&#9830</font>, 10<font color="red">&#9830</font>, J<font color="red">&#9830</font>, Q<font color="red">&#9830</font>, K<font color="red">&#9830</font>, 

A&spades;, 2&spades;, 3&spades;, 4&spades;, 5&spades;, 6&spades;, 7&spades;, 8&spades;, 9&spades;, 10&spades;, J&spades;, Q&spades;, K&spades;, 

A&clubs;, 2&clubs;, 3&clubs;, 4&clubs;, 5&clubs;, 6&clubs;, 7&clubs;, 8&clubs;, 9&clubs;, 10&clubs;, J&clubs;, Q&clubs;, K&clubs; 

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The probability of drawing a HEART:

There are 13 ways of succeeding by drawing one of these:

A<font color="red">&hearts;</font>, 2<font color="red">&hearts;</font>, 3<font color="red">&hearts;</font>, 4<font color="red">&hearts;</font>, 5<font color="red">&hearts;</font>, 6<font color="red">&hearts;</font>, 7<font color="red">&hearts;</font>, 8<font color="red">&hearts;</font>, 9<font color="red">&hearts;</font>, 10<font color="red">&hearts;</font>, J<font color="red">&hearts;</font>, Q<font color="red">&hearts;</font>, K<font color="red">&hearts;</font> 

There are 52 ways of either succeeding or failing, i.e., 
getting SOME card.

A<font color="red">&hearts;</font>, 2<font color="red">&hearts;</font>, 3<font color="red">&hearts;</font>, 4<font color="red">&hearts;</font>, 5<font color="red">&hearts;</font>, 6<font color="red">&hearts;</font>, 7<font color="red">&hearts;</font>, 8<font color="red">&hearts;</font>, 9<font color="red">&hearts;</font>, 10<font color="red">&hearts;</font>, J<font color="red">&hearts;</font>, Q<font color="red">&hearts;</font>, K<font color="red">&hearts;</font>,

A<font color="red">&#9830</font>, 2<font color="red">&#9830</font>, 3<font color="red">&#9830</font>, 4<font color="red">&#9830</font>, 5<font color="red">&#9830</font>, 6<font color="red">&#9830</font>, 7<font color="red">&#9830</font>, 8<font color="red">&#9830</font>, 9<font color="red">&#9830</font>, 10<font color="red">&#9830</font>, J<font color="red">&#9830</font>, Q<font color="red">&#9830</font>, K<font color="red">&#9830</font>, 

A&spades;, 2&spades;, 3&spades;, 4&spades;, 5&spades;, 6&spades;, 7&spades;, 8&spades;, 9&spades;, 10&spades;, J&spades;, Q&spades;, K&spades;, 

A&clubs;, 2&clubs;, 3&clubs;, 4&clubs;, 5&clubs;, 6&clubs;, 7&clubs;, 8&clubs;, 9&clubs;, 10&clubs;, J&clubs;, Q&clubs;, K&clubs; 

So the probability is {{{13/52}}} which reduces to {{{1/4}}} as a fraction,
{{{0.25}}} as a decimal, or {{{"25%"}}} as a percent

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The probability of drawing an ACE:

There are 4 ways of succeeding by drawing one of these:

A<font color="red">&hearts;</font>,

A<font color="red">&#9830</font>,

A&spades;, 

A&clubs;

There are 52 ways of either succeeding or failing, 
i.e., getting SOME card.

A<font color="red">&hearts;</font>, 2<font color="red">&hearts;</font>, 3<font color="red">&hearts;</font>, 4<font color="red">&hearts;</font>, 5<font color="red">&hearts;</font>, 6<font color="red">&hearts;</font>, 7<font color="red">&hearts;</font>, 8<font color="red">&hearts;</font>, 9<font color="red">&hearts;</font>, 10<font color="red">&hearts;</font>, J<font color="red">&hearts;</font>, Q<font color="red">&hearts;</font>, K<font color="red">&hearts;</font>,

A<font color="red">&#9830</font>, 2<font color="red">&#9830</font>, 3<font color="red">&#9830</font>, 4<font color="red">&#9830</font>, 5<font color="red">&#9830</font>, 6<font color="red">&#9830</font>, 7<font color="red">&#9830</font>, 8<font color="red">&#9830</font>, 9<font color="red">&#9830</font>, 10<font color="red">&#9830</font>, J<font color="red">&#9830</font>, Q<font color="red">&#9830</font>, K<font color="red">&#9830</font>, 

A&spades;, 2&spades;, 3&spades;, 4&spades;, 5&spades;, 6&spades;, 7&spades;, 8&spades;, 9&spades;, 10&spades;, J&spades;, Q&spades;, K&spades;, 

A&clubs;, 2&clubs;, 3&clubs;, 4&clubs;, 5&clubs;, 6&clubs;, 7&clubs;, 8&clubs;, 9&clubs;, 10&clubs;, J&clubs;, Q&clubs;, K&clubs; 

So the probability is {{{4/52}}} which reduces to {{{1/13}}} as a fraction.

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The probability of drawing the ACE of HEARTS:

There is 1 way of succeeding, by drawing the ACE of HEARTS:

A<font color="red">&hearts;</font>

There are 52 ways of either succeeding or failing, 
i.e., getting SOME card.

A<font color="red">&hearts;</font>, 2<font color="red">&hearts;</font>, 3<font color="red">&hearts;</font>, 4<font color="red">&hearts;</font>, 5<font color="red">&hearts;</font>, 6<font color="red">&hearts;</font>, 7<font color="red">&hearts;</font>, 8<font color="red">&hearts;</font>, 9<font color="red">&hearts;</font>, 10<font color="red">&hearts;</font>, J<font color="red">&hearts;</font>, Q<font color="red">&hearts;</font>, K<font color="red">&hearts;</font>,

A<font color="red">&#9830</font>, 2<font color="red">&#9830</font>, 3<font color="red">&#9830</font>, 4<font color="red">&#9830</font>, 5<font color="red">&#9830</font>, 6<font color="red">&#9830</font>, 7<font color="red">&#9830</font>, 8<font color="red">&#9830</font>, 9<font color="red">&#9830</font>, 10<font color="red">&#9830</font>, J<font color="red">&#9830</font>, Q<font color="red">&#9830</font>, K<font color="red">&#9830</font>, 

A&spades;, 2&spades;, 3&spades;, 4&spades;, 5&spades;, 6&spades;, 7&spades;, 8&spades;, 9&spades;, 10&spades;, J&spades;, Q&spades;, K&spades;, 

A&clubs;, 2&clubs;, 3&clubs;, 4&clubs;, 5&clubs;, 6&clubs;, 7&clubs;, 8&clubs;, 9&clubs;, 10&clubs;, J&clubs;, Q&clubs;, K&clubs; 

So the probability is {{{1/52}}}.

Edwin</pre>