What Is The Probability Of Drawing All Four Suits In Four Draws
Playing Cards Probability
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.eastward. spades♠hearts♥, diamonds♦, clubs♣.
Cards of Spades and clubs are blackness cards.
Cards of hearts and diamonds are crimson cards.
The card in each adjust, are ace, male monarch, queen, jack or knaves, 10, nine, viii, 7, 6, 5, four, 3 and 2.
King, Queen and Jack (or Knaves) are face cards. Then, there are 12 confront cards in the deck of 52 playing cards.
Worked-out problems on Playing cards probability:
one. A carte du jour 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 ruby-red colour
(iv) a card of diamond
(v) a king or a queen
(six) a non-confront card
(vii) a black face card
(viii) a black card
(ix) a not-ace
(x) non-face up carte du jour of black colour
(11) neither a spade nor a jack
(xii) neither a middle nor a red king
Solution:
In a playing card at that place 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 consequence
= 1/52
(ii) a jack
Number of favourable outcomes i.e. 'a jack' is four out of 52 cards.
Therefore, probability of getting 'a jack'
Number of favorable outcomes
P(B) = Full number of possible outcome
= iv/52
= 1/13
(iii) a king of red color
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 issue
= ii/52
= 1/26
(iv) a card of diamond
Number of favourable outcomes i.e. 'a menu of diamond' is xiii out of 52 cards.
Therefore, probability of getting 'a carte of diamond'
Number of favorable outcomes
P(D) = Full number of possible outcome
= thirteen/52
= 1/4
(5) a king or a queen
Full number of king is iv out of 52 cards.
Total number of queen is 4 out of 52 cards
Number of favourable outcomes i.e. 'a male monarch 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(East) = Total number of possible issue
= 8/52
= 2/13
(vi) a not-face card
Total number of face up carte du jour out of 52 cards = iii times 4 = 12
Total number of not-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 result
= 40/52
= 10/13
(vii) a blackness face carte du jour:
Cards of Spades and Clubs are black cards.
Number of confront card in spades (king, queen and jack or knaves) = iii
Number of face up card in clubs (king, queen and jack or knaves) = 3
Therefore, full number of blackness face up card out of 52 cards = 3 + 3 = half-dozen
Therefore, probability of getting 'a black face card'
Number of favorable outcomes
P(G) = Full number of possible effect
= 6/52
= 3/26
(viii) a black card:
Cards of spades and clubs are blackness cards.
Number of spades = 13
Number of clubs = 13
Therefore, full number of blackness card out of 52 cards = 13 + 13 = 26
Therefore, probability of getting 'a black card'
Number of favorable outcomes
P(H) = Full number of possible outcome
= 26/52
= one/2
(ix) a non-ace:
Number of ace cards in each of four suits namely spades, hearts, diamonds and clubs = one
Therefore, full number of ace cards out of 52 cards = four
Thus, full number of not-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 carte du jour of black color:
Cards of spades and clubs are black cards.
Number of spades = xiii
Number of clubs = thirteen
Therefore, full number of black card out of 52 cards = 13 + xiii = 26
Number of face cards in each suits namely spades and clubs = 3 + 3 = half-dozen
Therefore, total number of non-face carte du jour of black colour out of 52 cards = 26 - 6 = 20
Therefore, probability of getting 'non-face up bill of fare of black color'
Number of favorable outcomes
P(J) = Full number of possible outcome
= twenty/52
= v/thirteen
(xi) neither a spade nor a jack
Number of spades = 13
Total number of non-spades out of 52 cards = 52 - thirteen = 39
Number of jack out of 52 cards = iv
Number of jack in each of 3 suits namely hearts, diamonds and clubs = 3
[Since, 1 jack is already included in the xiii spades so, here we will accept number of jacks is iii]
Neither a spade nor a jack = 39 - three = 36
Therefore, probability of getting 'neither a spade nor a jack'
Number of favorable outcomes
P(1000) = Total number of possible outcome
= 36/52
= 9/13
(xii) neither a heart nor a cherry king
Number of hearts = xiii
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 scarlet cards.
Number of red kings in red cards = ii
Therefore, neither a heart nor a crimson male monarch = 39 - 1 = 38
[Since, i cerise king is already included in the thirteen hearts so, here we will have number of ruby-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 issue
= 38/52
= xix/26
two. A carte is fatigued at random from a well-shuffled pack of cards numbered 1 to 20. Find the probability of
(i) getting a number less than 7
(2) getting a number divisible by 3.
Solution:
(i) Total number of possible outcomes = twenty ( since there are cards numbered 1, 2, 3, ..., 20).
Number of favourable outcomes for the event E
= number of cards showing less than seven = 6 (namely 1, 2, 3, 4, 5, 6).
So, P(Due east) = \(\frac{\textrm{Number of Favourable Outcomes for the Event E}}{\textrm{Full Number of Possible Outcomes}}\)
= \(\frac{half-dozen}{20}\)
= \(\frac{3}{x}\).
(ii) Total number of possible outcomes = twenty.
Number of favourable outcomes for the consequence F
= number of cards showing a number divisible past 3 = 6 (namely 3, six, nine, 12, 15, 18).
And so, P(F) = \(\frac{\textrm{Number of Favourable Outcomes for the Event F}}{\textrm{Full Number of Possible Outcomes}}\)
= \(\frac{6}{twenty}\)
= \(\frac{3}{10}\).
3. A card is drawn at random from a pack of 52 playing cards. Find the probability that the menu drawn is
(i) a rex
(two) neither a queen nor a jack.
Solution:
Full number of possible outcomes = 52 (As there are 52 different cards).
(i) Number of favourable outcomes for the issue Eastward = number of kings in the pack = 4.
So, by definition, P(E) = \(\frac{4}{52}\)
= \(\frac{i}{13}\).
(ii) Number of favourable outcomes for the consequence F
= number of cards which are neither a queen nor a jack
= 52 - four - 4, [Since there are 4 queens and iv jacks].
= 44
Therefore, past definition, P(F) = \(\frac{44}{52}\)
= \(\frac{11}{xiii}\).
These are the basic problems on probability with playing cards.
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