How Many Ways Can You Draw 2 Red Cards
Playing Cards Probability
Playing cards probability issues based on a well-shuffled deck of 52 cards.
Basic concept on cartoon a card:
In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i.east. spades♠hearts♥, diamonds♦, clubs♣.
Cards of Spades and clubs are black cards.
Cards of hearts and diamonds are cherry-red cards.
The card in each suit, are ace, king, queen, jack or knaves, 10, nine, 8, vii, half-dozen, 5, 4, 3 and 2.
Rex, Queen and Jack (or Knaves) are face up cards. So, in that location are 12 face cards in the deck of 52 playing cards.
Worked-out problems on Playing cards probability:
1. A carte du jour is drawn from a well shuffled pack of 52 cards. Discover the probability of:
(i) 'two' of spades
(two) a jack
(iii) a king of red colour
(iv) a card of diamond
(v) a king or a queen
(half-dozen) a non-face menu
(seven) a black face up carte du jour
(viii) a black card
(ix) a non-ace
(ten) not-face carte du jour of black color
(11) neither a spade nor a jack
(xii) neither a heart nor a cherry-red king
Solution:
In a playing card at that place are 52 cards.
Therefore the full number of possible outcomes = 52
(i) '2' of spades:
Number of favourable outcomes i.e. 'ii' of spades is ane out of 52 cards.
Therefore, probability of getting '2' of spade
Number of favorable outcomes
P(A) = Full number of possible outcome
= 1/52
(ii) a jack
Number of favourable outcomes i.due east. 'a jack' is 4 out of 52 cards.
Therefore, probability of getting 'a jack'
Number of favorable outcomes
P(B) = Full number of possible outcome
= 4/52
= 1/13
(iii) a king of red colour
Number of favourable outcomes i.e. 'a rex of scarlet colour' is 2 out of 52 cards.
Therefore, probability of getting 'a king of cherry-red colour'
Number of favorable outcomes
P(C) = Total number of possible effect
= 2/52
= 1/26
(4) a card of diamond
Number of favourable outcomes i.e. 'a carte of diamond' is thirteen 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 male monarch or a queen' is 4 + iv = eight out of 52 cards.
Therefore, probability of getting 'a king or a queen'
Number of favorable outcomes
P(E) = Full number of possible upshot
= 8/52
= 2/13
(vi) a non-face bill of fare
Full number of face up card out of 52 cards = iii times 4 = 12
Total number of non-face menu out of 52 cards = 52 - 12 = 40
Therefore, probability of getting 'a not-face card'
Number of favorable outcomes
P(F) = Total number of possible outcome
= 40/52
= 10/thirteen
(vii) a black confront bill of fare:
Cards of Spades and Clubs are black cards.
Number of face carte du jour in spades (king, queen and jack or knaves) = iii
Number of confront card in clubs (male monarch, queen and jack or knaves) = 3
Therefore, full number of black face card out of 52 cards = 3 + 3 = six
Therefore, probability of getting 'a black face card'
Number of favorable outcomes
P(G) = Total number of possible effect
= half dozen/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 black carte out of 52 cards = 13 + thirteen = 26
Therefore, probability of getting 'a black carte'
Number of favorable outcomes
P(H) = Total number of possible outcome
= 26/52
= 1/ii
(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 = iv
Thus, full number of non-ace cards out of 52 cards = 52 - four
= 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-confront carte of black colour:
Cards of spades and clubs are black cards.
Number of spades = 13
Number of clubs = thirteen
Therefore, total number of black menu out of 52 cards = xiii + 13 = 26
Number of confront cards in each suits namely spades and clubs = iii + three = 6
Therefore, total number of not-face menu of black colour out of 52 cards = 26 - six = 20
Therefore, probability of getting 'not-face menu of black colour'
Number of favorable outcomes
P(J) = Total number of possible outcome
= 20/52
= 5/xiii
(eleven) neither a spade nor a jack
Number of spades = thirteen
Full number of not-spades out of 52 cards = 52 - 13 = 39
Number of jack out of 52 cards = 4
Number of jack in each of 3 suits namely hearts, diamonds and clubs = three
[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 - iii = 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 = 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 blood-red cards.
Number of ruddy kings in red cards = 2
Therefore, neither a heart nor a red king = 39 - i = 38
[Since, 1 red king is already included in the thirteen hearts so, here we will take number of ruby-red kings is 1]
Therefore, probability of getting 'neither a heart nor a blood-red rex'
Number of favorable outcomes
P(Fifty) = Full number of possible consequence
= 38/52
= 19/26
2. A card is drawn at random from a well-shuffled pack of cards numbered 1 to xx. Find the probability of
(i) getting a number less than vii
(two) getting a number divisible by 3.
Solution:
(i) Total number of possible outcomes = xx ( since at that place are cards numbered 1, 2, 3, ..., 20).
Number of favourable outcomes for the issue Due east
= number of cards showing less than 7 = half dozen (namely 1, 2, three, 4, v, half-dozen).
So, P(E) = \(\frac{\textrm{Number of Favourable Outcomes for the Event E}}{\textrm{Total Number of Possible Outcomes}}\)
= \(\frac{6}{20}\)
= \(\frac{3}{ten}\).
(ii) Full number of possible outcomes = xx.
Number of favourable outcomes for the event F
= number of cards showing a number divisible by three = 6 (namely 3, 6, ix, 12, 15, 18).
Then, P(F) = \(\frac{\textrm{Number of Favourable Outcomes for the Event F}}{\textrm{Total Number of Possible Outcomes}}\)
= \(\frac{6}{20}\)
= \(\frac{iii}{10}\).
3. A card is drawn at random from a pack of 52 playing cards. Detect the probability that the menu drawn is
(i) a king
(2) neither a queen nor a jack.
Solution:
Full number of possible outcomes = 52 (As there are 52 unlike cards).
(i) Number of favourable outcomes for the event Due east = 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 upshot F
= number of cards which are neither a queen nor a jack
= 52 - 4 - 4, [Since there are four queens and 4 jacks].
= 44
Therefore, past definition, P(F) = \(\frac{44}{52}\)
= \(\frac{11}{thirteen}\).
These are the basic problems on probability with playing cards.
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