Question

Find the probability of drawing an Ace or a Spade from a well shuffled pack of 52 playing cards.

Answer 1

Hint: We can draw an Ace or a Spade from a well shuffled pack of 52 playing cards, the number of ace cards present are 4 and spade cards are 13. These ace cards belong to each of the four suites including spade and thus we will subtract its probability (of intersection of the two). By applying the formula of probability and using the identity of union, we can calculate the value of required probability.
Formula to be used:
 $ P = \dfrac{f}{T} $ where, P is the probability, f , favorable outcomes and T, the total outcomes.
Identity to be used:
 $ P\left( {A \cup S} \right) = P(A) + P(S) - P\left( {A \cap S} \right) $
Where, $ \cap , \cup $ sign denotes intersection and union respectively, A and S denotes different events.

Complete step-by-step answer:
In a well shuffled deck of 52 playing cards, the number of spades and aces are:
Spade cards = 13
Ace cards = 4
The respective probabilities are given as:
Spade:
 $ P = \dfrac{f}{T} $ , here,
Favorable outcomes (f) = 13
Total outcomes (T) = 52
 $ \Rightarrow P\left( S \right) = \dfrac{{13}}{{52}} $
Ace:
 $ P = \dfrac{f}{T} $ , here,
Favorable outcomes (f) = 4
Total outcomes (T) = 52
 $ \Rightarrow P\left( S \right) = \dfrac{4}{{52}} $
Then, $ P\left( {A \cap S} \right) $ can be given as:
We have only 1 ace card belonging to spade suite, so the favorable outcome is 1
 $ \Rightarrow P\left( {A \cap S} \right) = \dfrac{1}{{52}} $
Venn diagram for the probability drawing either spade or ace is:
In the union of probability of both the cards, we don’t require the area of their intersection denoting selection of ace that belongs to spade suite. So to find the required probability, we can use the formula:
 $ P\left( {A \cup S} \right) = P(A) + P(S) - P\left( {A \cap S} \right) $
Substituting the values, we get:
\[
  P\left( {A \cup S} \right) = \dfrac{4}{{52}} + \dfrac{{13}}{{52}} - \dfrac{1}{{52}} \\
  P\left( {A \cup S} \right) = \dfrac{{16}}{{52}} \\
  P\left( {A \cup S} \right) = \dfrac{4}{{13}} \;
 \]
Therefore, the probability of drawing an Ace or a Spade from a well shuffled pack of 52 playing cards is $ \dfrac{4}{{13}} $
So, the correct answer is “ $ \dfrac{4}{{13}} $ ”.

Note: The question can also be solved by the following method:
We know that ace cards are 4 in number belonging to each suit and spade cards are 13 in number including 1 ace of spade.
So total number of ace and spade cards are:
\[4 + {\text{1}}3 = 17\]
But, we can either draw a spade card or an ace card, so for the favorable outcomes, we need to subtract this 1 from their addition
 \[17-1 = 16\]
Applying the formula, we get:
 $
  P = \dfrac{{16}}{{52}} \\
  P = \dfrac{4}{{13}} \;
  $