Question

From a normal pack of \[52\] cards, one card is selected at random. How can I find the probability of the card being either a black card or an ace?

Answer 1

Hint: To solve this question, we will first consider two events, one of drawing an ace card and another of drawing a black card. Then, we will find the probability of these events individually. Then we will find the probability of the union of these events using the simple formula:
\[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\]

Complete answer:
Let, A be the event of drawing an ace and B be the event of drawing a black coloured card.
So, what we have to find is \[P\left( {A \cup B} \right)\].
Now we know that there are four aces in a pack. So, one can be chosen in four ways.
Also, the total cards are \[52\].
\[\therefore P\left( A \right) = \dfrac{4}{{52}}\]
\[ \Rightarrow P\left( A \right) = \dfrac{1}{{13}}\]
Now we know that out of \[52\] cards there are a total \[26\] black cards. So, the probability of event B is;
\[ \Rightarrow P\left( B \right) = \dfrac{{26}}{{52}}\]
\[ \Rightarrow P\left( B \right) = \dfrac{1}{2}\]
Now to find \[P\left( {A \cup B} \right)\], we must be knowing \[P\left( {A \cap B} \right)\].
Now \[P\left( {A \cap B} \right)\] means an ace card that is black in colour. Now, from the total card there are two black-coloured aces. So,
\[P\left( {A \cap B} \right) = \dfrac{2}{{52}}\]
\[ \Rightarrow P\left( {A \cap B} \right) = \dfrac{1}{{26}}\]
Now we will put these values in the formula;
\[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\]
So, we have;
\[ \Rightarrow P\left( {A \cup B} \right) = \dfrac{1}{{13}} + \dfrac{1}{2} - \dfrac{1}{{26}}\]
On taking the LCM we have;
\[ \Rightarrow P\left( {A \cup B} \right) = \dfrac{{2 + 13 - 1}}{{26}}\]
On solving we get;
\[ \Rightarrow P\left( {A \cup B} \right) = \dfrac{{14}}{{26}}\]
On reducing the RHS we get;
\[ \Rightarrow P\left( {A \cup B} \right) = \dfrac{7}{{13}}\]
Hence the required probability is \[\dfrac{7}{{13}}\].

Note: One important thing to note is that the probability of an event lies between zero and one. It cannot be more than one. If an event has probability one then, it means that it is a certain event. The probability of non-occurrence of an event is equal to one minus the probability of its occurrence.