What Is The Probability Of Drawing A Red Card And Then Spinning A 1?
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.e. spades♠hearts♥, diamonds♦, clubs♣.
Cards of Spades and clubs are blackness cards.
Cards of hearts and diamonds are carmine cards.
The carte in each suit, are ace, king, queen, jack or knaves, 10, 9, viii, 7, half-dozen, five, 4, 3 and two.
King, Queen and Jack (or Knaves) are face up cards. So, there are 12 face up cards in the deck of 52 playing cards.
Worked-out problems on Playing cards probability:
1. A card is fatigued 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
(half dozen) a non-confront card
(vii) a black face card
(viii) a blackness bill of fare
(nine) a non-ace
(x) non-face carte du jour of blackness colour
(xi) neither a spade nor a jack
(xii) neither a eye nor a red king
Solution:
In a playing bill of fare there are 52 cards.
Therefore the full number of possible outcomes = 52
(i) '2' of spades:
Number of favourable outcomes i.due east. '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.eastward. 'a jack' is iv out of 52 cards.
Therefore, probability of getting 'a jack'
Number of favorable outcomes
P(B) = Total number of possible outcome
= 4/52
= i/13
(iii) a king of carmine colour
Number of favourable outcomes i.e. 'a king of red colour' is 2 out of 52 cards.
Therefore, probability of getting 'a rex of blood-red colour'
Number of favorable outcomes
P(C) = Total number of possible event
= ii/52
= 1/26
(4) a bill of fare of diamond
Number of favourable outcomes i.east. '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 result
= 13/52
= i/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 + four = 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
= eight/52
= 2/13
(vi) a non-face card
Full number of face up bill of fare out of 52 cards = iii times iv = 12
Total number of not-face bill of fare out of 52 cards = 52 - 12 = forty
Therefore, probability of getting 'a non-confront card'
Number of favorable outcomes
P(F) = Total number of possible outcome
= xl/52
= ten/xiii
(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) = iii
Number of face up carte in clubs (king, queen and jack or knaves) = 3
Therefore, full number of blackness confront card out of 52 cards = three + 3 = 6
Therefore, probability of getting 'a black confront card'
Number of favorable outcomes
P(Grand) = 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, full number of black card out of 52 cards = 13 + 13 = 26
Therefore, probability of getting 'a blackness card'
Number of favorable outcomes
P(H) = Total number of possible issue
= 26/52
= 1/2
(nine) a not-ace:
Number of ace cards in each of iv 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 not-ace'
Number of favorable outcomes
P(I) = Total number of possible upshot
= 48/52
= 12/13
(x) not-confront carte of blackness colour:
Cards of spades and clubs are blackness cards.
Number of spades = xiii
Number of clubs = thirteen
Therefore, total number of blackness bill of fare out of 52 cards = thirteen + 13 = 26
Number of confront cards in each suits namely spades and clubs = iii + three = 6
Therefore, total number of non-face card of black color out of 52 cards = 26 - vi = xx
Therefore, probability of getting 'non-face card of black color'
Number of favorable outcomes
P(J) = Total number of possible upshot
= xx/52
= five/13
(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 = 4
Number of jack in each of iii suits namely hearts, diamonds and clubs = 3
[Since, ane jack is already included in the thirteen spades so, here we will accept 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 male monarch
Number of hearts = thirteen
Total number of non-hearts out of 52 cards = 52 - thirteen = 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 - i = 38
[Since, 1 red male monarch is already included in the 13 hearts then, here we will have number of crimson kings is one]
Therefore, probability of getting 'neither a heart nor a red rex'
Number of favorable outcomes
P(L) = Full number of possible outcome
= 38/52
= 19/26
2. A menu is drawn at random from a well-shuffled pack of cards numbered ane to 20. Find the probability of
(i) getting a number less than vii
(two) getting a number divisible by three.
Solution:
(i) Total number of possible outcomes = 20 ( since there are cards numbered i, 2, three, ..., 20).
Number of favourable outcomes for the event E
= number of cards showing less than vii = 6 (namely one, 2, 3, 4, 5, 6).
So, P(E) = \(\frac{\textrm{Number of Favourable Outcomes for the Event Due east}}{\textrm{Total Number of Possible Outcomes}}\)
= \(\frac{6}{20}\)
= \(\frac{3}{ten}\).
(ii) Total number of possible outcomes = xx.
Number of favourable outcomes for the event F
= number of cards showing a number divisible by 3 = half-dozen (namely 3, six, 9, 12, 15, xviii).
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 fatigued 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 (Every bit there are 52 different cards).
(i) Number of favourable outcomes for the event E = number of kings in the pack = 4.
And then, by definition, P(E) = \(\frac{4}{52}\)
= \(\frac{ane}{thirteen}\).
(ii) Number of favourable outcomes for the event F
= number of cards which are neither a queen nor a jack
= 52 - 4 - 4, [Since in that location are 4 queens and iv jacks].
= 44
Therefore, by definition, P(F) = \(\frac{44}{52}\)
= \(\frac{11}{13}\).
These are the basic issues on probability with playing cards.
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