Answer

Verified

393.6k+ views

**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}} \;

$

Recently Updated Pages

The base of a right prism is a pentagon whose sides class 10 maths CBSE

A die is thrown Find the probability that the number class 10 maths CBSE

A mans age is six times the age of his son In six years class 10 maths CBSE

A started a business with Rs 21000 and is joined afterwards class 10 maths CBSE

Aasifbhai bought a refrigerator at Rs 10000 After some class 10 maths CBSE

Give a brief history of the mathematician Pythagoras class 10 maths CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Name 10 Living and Non living things class 9 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Write the 6 fundamental rights of India and explain in detail