Suppose a card is drawn at random from a deck of 52 cards what is the probability of getting a 10

Suppose a card is drawn at random from a deck of 52 cards what is the probability of getting a 10

Important Notes

  • The sample space for a set of cards is 52 as there are 52 cards in a deck. This makes the denominator for finding the probability of drawing a card as 52.
  • Learn more about related terminology of probability to solve problems on card probability better. 

The suits which are represented by red cards are hearts and diamonds while the suits represented by black cards are spades and clubs.

There are 26 red cards and 26 black cards. 

Let's learn about the suits in a deck of cards.

Suits in a deck of cards are the representations of red and black color on the cards.

Based on suits, the types of cards in a deck are: 

There are 52 cards in a deck.

Each card can be categorized into 4 suits constituting 13 cards each.

These cards are also known as court cards.

They are Kings, Queens, and Jacks in all 4 suits.

All the cards from 2 to 10 in any suit are called the number cards. 

These cards have numbers on them along with each suit being equal to the number on number cards. 

There are 4 Aces in every deck, 1 of every suit. 

Suppose a card is drawn at random from a deck of 52 cards what is the probability of getting a 10

Tips and Tricks

  • There are 13 cards of each suit, consisting of 1 Ace, 3 face cards, and 9 number cards.
  • There are 4 Aces, 12 face cards, and 36 number cards in a 52 card deck.
  • Probability of drawing any card will always lie between 0 and 1.
  • The number of spades, hearts, diamonds, and clubs is same in every pack of 52 cards.

Now that you know all about facts about a deck of cards, you can draw a card from a deck and find its probability easily.

How to Determine the Probability of Drawing a Card?

Let's learn how to find probability first.

Now you know that probability is the ratio of number of favorable outcomes to the number of total outcomes, let's apply it here.

Examples

Example 1: What is the probability of drawing a king from a deck of cards?

Solution: Here the event E is drawing a king from a deck of cards.

There are 52 cards in a deck of cards. 

Hence, total number of outcomes = 52

The number of favorable outcomes = 4 (as there are 4 kings in a deck)

Hence, the probability of this event occuring is 

P(E) = 4/52 = 1/13

\(\therefore\) Probability of drawing a king from a deck of cards is 1/13.

Example 2: What is the probability of drawing a black card from a pack of cards?

Solution: Here the event E is drawing a black card from a pack of cards.

The total number of outcomes = 52

The number of favorable outcomes = 26

Hence, the probability of event occuring is 

P(E) = 26/52 = 1/2

\(\therefore\) Probability of drawing a black card from a pack of cards is 1/2.

Solved Examples

Jessica has drawn a card from a well-shuffled deck. Help her find the probability of the card either being red or a King.

Solution

Jessica knows here that event E is the card drawn being either red or a King.

The total number of outcomes = 52

There are 26 red cards, and 4 cards which are Kings.

However, 2 of the red cards are Kings.

If we add 26 and 4, we will be counting these two cards twice.

Thus, the correct number of outcomes which are favorable to E is

26 + 4 - 2 = 28

Hence, the probability of event occuring is

P(E) = 28/52 = 7/13

\(\therefore\) Probability of card either being red or a King card is 7/13.

Help Diane determine the probability of the following:

  • Drawing a Red Queen
  • Drawing a King of Spades
  • Drawing a Red Number Card 

Solution

Diane knows here the events E1, E2, and E3 are Drawing a Red Queen, Drawing a King of Spades, and Drawing a Red Number Card.

The total number of outcomes in every case = 52

There are 26 red cards, of which 2 are Queens.

Hence, the probability of event E1 occuring is

P(E1) = 2/52 = 1/26

There are 13 cards in each suit, of which 1 is King.

Hence, the probability of event E2 occuring is

P(E2) = 1/52 

  • Drawing a Red Number Card

There are 9 number cards in each suit and there are 2 suits which are red in color. 

There are 18 red number cards.

Hence, the probability of event E3 occuring is

P(E3) = 18/52 = 9/26 

\(\therefore\) Diane determined that the probabilities are P(E1) = 1/26, P(E2) = 1/52, and P(E3) = 9/26.

Interactive Questions

Here are a few activities for you to practice.

Select/Type your answer and click the "Check Answer" button to see the result.

We hope you enjoyed learning about probability of drawing a card from a pack of 52 cards with the practice questions. Now you will easily be able to solve problems on number of cards in a deck, face cards in a deck, 52 card deck, spades hearts diamonds clubs in pack of cards. Now you can draw a card from a deck and find its probability easily .

The mini-lesson targeted the fascinating concept of card probability. The math journey around card probability starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that is not only relatable and easy to grasp, but will also stay with them forever. Here lies the magic with Cuemath.

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At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students! Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic. Be it problems, online classes, videos, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

We find the ratio of the favorable outcomes as per the condition of drawing the card to the total number of outcomes, i.e, 52.

2. What is the probability of drawing any face card?

Probability of drawing any face card is 6/26.

3. What is the probability of drawing a red card?

Probability of drawing a red card is 1/2.

4. What is the probability of drawing a king or a red card?

Probability of drawing a king or a red card is 7/13.

5. What is the probability of drawing a king or a queen?

The probability of drawing a king or a queen is 2/13.

6. What are the 5 rules of probability?

The 5 rules of probability are:

For any event E, the probability of occurence of E will always lie between 0 and 1

The sum of probabilities of every possible outcome will always be 1

The sum of probability of occurence of E and probability of E not occuring will always be 1

When any two events are not disjoint, the probability of occurence of A and B is not 0 while when two events are disjoint, the probability of occurence of A and B is 0.

As per this rule, P(A or B) = (P(A) + P(B) - P(A and B)).

7. What is the probability of drawing a king of hearts?

Probability of drawing a king of hearts is 1/52.

8. Is Ace a face card in probability?

No, Ace is not a face card in probability.

9. What is the probability it is not a face card?

The probability it is not a face card is 10/13.

10. How many black non-face cards are there in a deck?

There are 20 black non-face cards in a deck.

Playing cards probability problems based on a well-shuffled deck of 52 cards.

Basic concept on drawing a card:

In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i.e. spades ♠ hearts , diamonds , clubs .

Cards of Spades and clubs are black cards.

Cards of hearts and diamonds are red cards.

The card in each suit, are ace, king, queen, jack or knaves, 10, 9, 8, 7, 6, 5, 4, 3 and 2.

King, Queen and Jack (or Knaves) are face cards. So, there are 12 face cards in the deck of 52 playing cards.


Worked-out problems on Playing cards probability:

1. A card is drawn from a well shuffled pack of 52 cards. Find the probability of:

(i) ‘2’ of spades

(ii) a jack

(iii) a king of red colour

(iv) a card of diamond

(v) a king or a queen

(vi) a non-face card

(vii) a black face card

(viii) a black card

(ix) a non-ace

(x) non-face card of black colour

(xi) neither a spade nor a jack

(xii) neither a heart nor a red king

Solution:

In a playing card there are 52 cards.

Therefore the total number of possible outcomes = 52

(i) ‘2’ of spades:

Number of favourable outcomes i.e. ‘2’ of spades is 1 out of 52 cards.

Therefore, probability of getting ‘2’ of spade

               Number of favorable outcomes
P(A) =     Total number of possible outcome       = 1/52

(ii) a jack

Number of favourable outcomes i.e. ‘a jack’ is 4 out of 52 cards.

Therefore, probability of getting ‘a jack’

               Number of favorable outcomes
P(B) =     Total number of possible outcome       = 4/52       = 1/13

(iii) a king of red colour

Number of favourable outcomes i.e. ‘a king of red colour’ is 2 out of 52 cards.

Therefore, probability of getting ‘a king of red colour’

               Number of favorable outcomes
P(C) =     Total number of possible outcome       = 2/52       = 1/26

(iv) a card of diamond

Number of favourable outcomes i.e. ‘a card of diamond’ is 13 out of 52 cards.

Therefore, probability of getting ‘a card of diamond’

               Number of favorable outcomes
P(D) =     Total number of possible outcome       = 13/52       = 1/4

(v) a king or a queen

Total number of king is 4 out of 52 cards.

Total number of queen is 4 out of 52 cards

Number of favourable outcomes i.e. ‘a king or a queen’ is 4 + 4 = 8 out of 52 cards.

Therefore, probability of getting ‘a king or a queen’

               Number of favorable outcomes
P(E) =     Total number of possible outcome       = 8/52       = 2/13

(vi) a non-face card

Total number of face card out of 52 cards = 3 times 4 = 12

Total number of non-face card out of 52 cards = 52 - 12 = 40

Therefore, probability of getting ‘a non-face card’

               Number of favorable outcomes
P(F) =     Total number of possible outcome       = 40/52       = 10/13

(vii) a black face card:

Cards of Spades and Clubs are black cards.

Number of face card in spades (king, queen and jack or knaves) = 3

Number of face card in clubs (king, queen and jack or knaves) = 3

Therefore, total number of black face card out of 52 cards = 3 + 3 = 6

Therefore, probability of getting ‘a black face card’

               Number of favorable outcomes
P(G) =     Total number of possible outcome       = 6/52       = 3/26

(viii) a black card:

Cards of spades and clubs are black cards.

Number of spades = 13

Number of clubs = 13

Therefore, total number of black card out of 52 cards = 13 + 13 = 26

Therefore, probability of getting ‘a black card’

               Number of favorable outcomes
P(H) =     Total number of possible outcome       = 26/52       = 1/2

(ix) a non-ace:

Number of ace cards in each of four suits namely spades, hearts, diamonds and clubs = 1

Therefore, total number of ace cards out of 52 cards = 4

Thus, total number of non-ace cards out of 52 cards = 52 - 4

= 48

Therefore, probability of getting ‘a non-ace’

               Number of favorable outcomes
P(I) =     Total number of possible outcome       = 48/52       = 12/13

(x) non-face card of black colour:

Cards of spades and clubs are black cards.

Number of spades = 13

Number of clubs = 13

Therefore, total number of black card out of 52 cards = 13 + 13 = 26

Number of face cards in each suits namely spades and clubs = 3 + 3 = 6

Therefore, total number of non-face card of black colour out of 52 cards = 26 - 6 = 20

Therefore, probability of getting ‘non-face card of black colour’

               Number of favorable outcomes
P(J) =     Total number of possible outcome       = 20/52       = 5/13

(xi) neither a spade nor a jack

Number of spades = 13

Total number of non-spades out of 52 cards = 52 - 13 = 39

Number of jack out of 52 cards = 4

Number of jack in each of three suits namely hearts, diamonds and clubs = 3

[Since, 1 jack is already included in the 13 spades so, here we will take number of jacks is 3]

Neither a spade nor a jack = 39 - 3 = 36

Therefore, probability of getting ‘neither a spade nor a jack’

               Number of favorable outcomes
P(K) =     Total number of possible outcome       = 36/52       = 9/13

(xii) neither a heart nor a red king

Number of hearts = 13

Total number of non-hearts out of 52 cards = 52 - 13 = 39

Therefore, spades, clubs and diamonds are the 39 cards.

Cards of hearts and diamonds are red cards.

Number of red kings in red cards = 2

Therefore, neither a heart nor a red king = 39 - 1 = 38

[Since, 1 red king is already included in the 13 hearts so, here we will take number of red kings is 1]

Therefore, probability of getting ‘neither a heart nor a red king’

               Number of favorable outcomes
P(L) =     Total number of possible outcome       = 38/52       = 19/26

Suppose a card is drawn at random from a deck of 52 cards what is the probability of getting a 10

2. A card is drawn at random from a well-shuffled pack of cards numbered 1 to 20. Find the probability of

(i) getting a number less than 7

(ii) getting a number divisible by 3.

Solution:

(i) Total number of possible outcomes = 20 ( since there are cards numbered 1, 2, 3, ..., 20).

Number of favourable outcomes for the event E

                                = number of cards showing less than 7 = 6 (namely 1, 2, 3, 4, 5, 6).

So, P(E) = \(\frac{\textrm{Number of Favourable Outcomes for the Event E}}{\textrm{Total Number of Possible Outcomes}}\)

             = \(\frac{6}{20}\)

             = \(\frac{3}{10}\).


(ii) Total number of possible outcomes = 20.

Number of favourable outcomes for the event F

                                = number of cards showing a number divisible by 3 = 6 (namely 3, 6, 9, 12, 15, 18).

So, P(F) = \(\frac{\textrm{Number of Favourable Outcomes for the Event F}}{\textrm{Total Number of Possible Outcomes}}\)

             = \(\frac{6}{20}\)

             = \(\frac{3}{10}\).


3. A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is 

(i) a king

(ii) neither a queen nor a jack.

Solution:

Total number of possible outcomes = 52 (As there are 52 different cards).

(i) Number of favourable outcomes for the event E = number of kings in the pack = 4.

So, by definition, P(E) = \(\frac{4}{52}\)

                                 = \(\frac{1}{13}\).


(ii) Number of favourable outcomes for the event F

                    = number of cards which are neither a queen nor a jack

                    = 52 - 4 - 4, [Since there are 4 queens and 4 jacks].

                    = 44

Therefore, by definition, P(F) = \(\frac{44}{52}\)

                                          = \(\frac{11}{13}\).

These are the basic problems on probability with playing cards.

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