Question

In an experiment of drawing a card at random from a pack, the event of getting a spade is denoted by \[A\] and getting a pictured card (King, Queen or Jack) is denoted by \[B\]. Find the probabilities of \[A,B,A \cap B,A \cup B\].

Answer 1

Hint: Here, in the question, we are given two events of drawing a card randomly from a pack. And we are asked to find the probabilities of the required events. We will understand each of the events separately and use the simple formula of calculating probability to get the desired result.
Formula used:
Probability of an event E is given by, \[P(E) = \dfrac{\text{No. of favorable outcomes}}{\text{No. of total outcomes}}\]

Complete step-by-step solution:
Given,
\[A\] is the event that the card drawn is a spade
\[B\] is the event that the card drawn is a pictured card.
We know that there are a total of 52 cards in a pack. Out of 52 cards, there are only \[13\] spades cards.
To calculate the probability of event\[A\],
Probability of \[A\], \[P\left( A \right)\]=\[\dfrac{\text{Total No. of spades}}{\text{Total no. of cards}}\]
\[ \Rightarrow P\left( A \right) = \dfrac{{13}}{{52}} \\
   \Rightarrow P\left( A \right) = \dfrac{1}{4} \]
Now, we also know that there are \[12\left( {3 \times 4} \right)\] pictured cards in a pack.
To calculate the probability of event \[B\],
Probability of \[B\], \[P\left( B \right) = \dfrac{\text{Total no. of pictured cards}}{\text{Total no. of cards}}\]
\[ \Rightarrow P\left( B \right) = \dfrac{{12}}{{52}} \\
   \Rightarrow P\left( B \right) = \dfrac{3}{{13}} \]
Now, before calculating the probability of \[A \cap B\], let us first understand the meaning of \[A \cap B\].
\[A \cap B\] denotes the event \[A\;and\;B\], which means that the card drawn is a pictured card of spades.
As we already know that, there are \[3\] such cards which are both pictured cards and are of spades.
Therefore, Probability of event \[A\;and\;B\],\[P\left( {A \cap B} \right) = \dfrac{\text{Total no. of spades pictures card}}{\text{Total no. of cards}}\]
\[ \Rightarrow P\left( {A \cap B} \right) = \dfrac{3}{{52}}\]
\[A \cup B\] denotes the event \[A\;or\;B\], which means that the card drawn is either a spade card or a pictured card.
There are \[13\] spades cards and \[12\] pictured cards in total. But there are \[3\] common cards which are pictured and spades. So total cards in set \[A \cup B\] will be \[13 + 12 - 3 = 22\]
Therefore, Probability of event \[A\;or\;B\], \[P\left( {A \cup B} \right) = \dfrac{\text{Total no. of possible outcomes}}{\text{Total no. of cards}}\]
\[ \Rightarrow P\left( {A \cup B} \right) = \dfrac{{22}}{{52}} \\
  \Rightarrow P\left( {A \cup B} \right) = \dfrac{{11}}{{26}} \]

Note:It is important to understand the concepts of union and intersection, otherwise we can make a mistake in taking the total number of possible outcomes. Any minor mistake can lead to a wrong answer at the end. Or if anyone is still confused, then he/she must check the final answers by putting the values to the formula stated as \[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\]