Often it is required to compute the probability of an event given that another event has occurred. For example, what is the probability that two cards drawn at random from a deck of playing cards will both be aces? It might seem that you could use the formula for the probability of two independent events and simply multiply 4/52 x 4/52 = 1/169. This would be incorrect, however, because the two events are not independent. If the first card drawn is an ace, then the probability that the second card is also an ace would be lower because there would only be three aces left in the deck. Once the first card chosen is an ace, the probability that the second card chosen is also an ace is called the conditional probability of drawing an ace. In this case, the "condition" is that the first card is an ace. Symbolically, we write this as: P(ace on second draw | an ace on the first draw) The vertical bar "|" is read as "given," so the above expression is short for: "The probability that an ace is drawn on the second draw given that an ace was drawn on the first draw". What is this probability? Since after an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the probability that one of these aces will be drawn is 3/51 = 1/17. If Events A and B are not independent, then . Applying this to the problem of two aces, the probability of drawing two aces from a deck is 4/52 x 3/51 = 1/221. One more example: If you draw two cards from a deck, what is the probability that you will get the Ace of Diamonds and a black card? There are two ways you can satisfy this condition: (1) You can get the Ace of Diamonds first and then a black card or (2) you can get a black card first and then the Ace of Diamonds. Let's calculate Case A. The probability that the first card is the Ace of Diamonds is 1/52. The probability that the second card is black given that the first card is the Ace of Diamonds is 26/51 because 26 of the remaining 51 cards are black. The probability is therefore 1/52 x 26/51 = 1/102. Now for Case 2: the probability that the first card is black is 26/52 = 1/2. The probability that the second card is the Ace of Diamonds given that the first card is black is 1/51. The probability of Case 2 is therefore 1/2 x 1/51 = 1/102, the same as the probability of Case 1. Recall that the probability of or is . In this problem, since a card cannot be the Ace of Diamonds and be a black card. Therefore, the probability of Case 1 or Case 2 is 1/102 + 1/102 = 2/102 = 1/51. So, 1/51 is the probability that you will get the Ace of Diamonds and a black card when drawing two cards from 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 outcomesP(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 outcomesP(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 outcomesP(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 outcomesP(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 outcomesP(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 outcomesP(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 outcomesP(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 outcomesP(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 outcomesP(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 outcomesP(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 outcomesP(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 outcomesP(L) = Total number of possible outcome = 38/52 = 19/26 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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