There are two ways to think of this question, "Student" attempted to answer the first one I'll go over. "Student" is looking at the question as "If you draw a card from a deck of 52 cards, what is the probability that the next card you draw will have a matching value (not suit)." Since the deck is now only 51 cards, and there are only 3 more matching values since 1 has been removed, the answer of 3/51 is correct. Now, a slightly different question would be "If you draw two cards from a deck of 52 cards, what is the probability that the cards will be a pair (matching values)." This is a little more complex. There are 52 * 51 combinations of cards in a deck, you can imagine that for every one of the 52 cards, there are 51 potential matches. This is 2,652. There are 4 suits of 13 cards, and for every card, there are 3 other cards that can form a pair with it. If you draw an ace, there are 3 other aces to draw. Lets consider this A1, A2, A3, and A4. Drawing A1 and A4 is the same as A4 and A1, so this is a combination problem rather than a permutation problem, so the question (for just aces) is, how many ways are there of drawing an ace? The answer is 6: A1,A2 A1,A3 A1,A4 A2,A3 A2,A4 A3,A4 So there are 6 possibilities per value, 13 values total; 6 * 13 is 78, or written out, there are 78 ways to draw two cards out the deck that will be pairs. We know from above that there are 2,652 possible ways to draw two cards out of the deck, so the probability that a randomly drawn set of two cards is a pair is 78/2,652, or 0.0294, or 2.94%.
Chris P. a. Both are Diamonds. b. Neither is a Diamond. c. Both are of the same unit. d. Both are of the same suit given that the first card drawn is a Club. e. They are not of the same suit. f. They are not of the same suit, given that the first card drawn is a Club. X
1 Expert Answer
Kirill Z. answered • 09/29/13
Physics, math tutor with great knowledge and teaching skills
Since there is no replacement, the probability of the second draw depends on the first outcome. Let me illustrate on the part a.
a. There are 13 diamonds in the deck (just as any other suit). Since there are 52 cards, the probability of getting a diamond in the first draw is 13/52. After the first card is drawn, there are just 12 diamonds left. So, the probability of drawing the diamond now is 12/51 (remember, there is no replacement, so there are just 51 cards left after the first card is drawn!). The total probability is the product of two probabilities: the probability to draw diamond from original deck, which is 13/52 times the probability to draw a diamond from the remaining 51 cards, provided the first card drawing is diamond, which is 12/51.
So, p=13/52*12/51=1/4*4/17=1/17; Answer to a. part is 1/17.
Now you shall be able to answer part b. by analogy. I will give just the answer: 19/34. Let us consider part c now.
There are 13 units in the deck, with 4 cards of the same unit per each unit. So, drawing a given unit has a probability of 4/52=1/13. Once a card of a certain unit is drawn, there are just 3 cards of the same unit left, so the probability of drawing the second of the same unit is 3/51=1/17. The product of two probabilities is the total probability to draw two cards of the same given unit, that is 1/(13*17)=1/221. Since there are 13 units, this needs to be multiplied by the number of units, that is 13. Thus the answer is 1/221*13=1/17.
Now to part d. If the first card is Club, then there are 12 Clubs left and overall 51 cards left. Probability to draw a Club second time is 12/51=4/17.
Part e. Let us consider the probability that they are of the same suit. In part a. we determined the probability to have two diamonds. It is 1/17. So is the probability to get two cards of any other suit. Thus the total probability to get two cards of the same suit is 4*1/17=4/17. The probability that two cards are not of the same suit is just 1-4/17=13/17. Answer: 13/17.
Part f. If the first card is a Club, then the second MUST be NOT Club. There are 12 Clubs left and total 51 cards left. Thus there are 51-12=39 cards, that are NOT Clubs, left. Thus the probability to draw the second card, which is not a Club, is 39/51=13/17. Answer: 13/17.
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2 cards are drawn at random from the a deck of 52 (without replacement). Ace = 1, Jack = 11, Queen = 12, King = 13. How would you find the probability of getting 2 cards of consecutive numbers?
Note: It is not a cyclic so 1 is not a consecutive of King
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It is cyclic so: ...Jack, Q, K, A, 2...
The first card you pick doesn't matter. You have a 100% chance of drawing a card that fits the requirements. There are now 51 cards in the deck. If the card was the n of some suit, the next card must be (n+1) or (n-1) of that same suit. Therefore there are two cards in the 51 that will satisfy the requirements. So, the P of drawing two consecutive cards of the same suit is 2/51
I wasn't given an answer, I was simply told I was wrong. Where is my error?