Find the probability of getting an ace from a well shuffled deck of 52 playing cards?

Suppose that a deck of 52 cards containing four aces is shuffled thoroughly and the cards are then distributed among four players so that each player receives 13 cards. Determine the probability that each player will receive one ace. The answer to this is given as$$\frac$ since you are choosing 13 cards for 4 people. There are $$\binom$$ $\begingroup$ If you are referring to the text's answer, notice that we are choosing positions for the aces in both the numerator and the denominator. In the numerator, we are choosing the position for the ace in each person's hand. In the denominator, we are choosing the position of the aces in the deck. $\endgroup$ All we really care about is the placement for the aces; not where the other cards may be in the deck, nor even the suits of the aces. So let us take 52 blank cards and 4 stickers with 'ace' written on them. To 'deal the cards', place the blank cards in for lines of 13, then unbiasedly select four from them and put a sticker on each. How may ways are there to stick the aces on 4 different cards in the deck of 52? That is a selection of 4 from 52. $$\binom $$ Informal analysis: we can re-imagine the process of the deal as one in which there are 52 slots, 13 assigned to each hand and the cards are distributed from the top unifomly to the remaining slots. We put the four aces on the top of the deck. The first ace can be dealt to any of the four players. The second ace can be dealt to any of the remaining 39 slots. The third ace ca...

A card is drawn at random from a pack of 5 2 cards. Find the probability that the card is drawn is (i) a black king (ii) either a black card or a king (iii) a jack, queen or a king (iv) neither an ace nor a king (v) spade or an ace (vi) neither a red card nor a queen (vii) other than an ace (viii) a ten (ix) a spade (x) a black card (xi) the seven of clubs (xii) jack (xiii) the ace of spades (xiv) a queen (xv) a heart (xvi) a red card (xvii) neither a king nor a queen

It is not indicated whether the first card is replaced before the second is taken... Let's consider both cases. #Prob = ("number of desirable outcomes")/("total number of possible outcomes")# There are 4 aces in a deck of cards which has 52 cards in total. If the first card is NOT replaced: P (Ace, Ace) = #4/52 xx 3/51 = 1/221 # (The number of aces remaining is 1 less, and there is 1 less card to choose from.) If the first card IS replaced , then the probability of drawing an ace stays the same every time a card is chosen: P (Ace, Ace) = #4/52 xx 4/52 = 1/169# Note: A probability should be given as a fraction, in the simplest form. It is always a value from 0 to 1 0 #rArr# (the outcome is impossible) 1 #rArr# (the outcome is guaranteed)

• • • • Question:30. Suppose that you draw two cards from a well-shuffled deck of 52 playing cards without replacement. What is the probability that the second card is an ace, given that the first card is an ace? 32. Suppose that you draw two cards from a well-shuffled deck of 52 playing cards without replacement. What is the probability that the second card is red, 30. Suppose that you draw two cards from a well-shuffled deck of 52 playing cards without replacement. What is the probability that the second card is an ace, given that the first card is an ace? 32. Suppose that you draw two cards from a well-shuffled deck of 52 playing cards without replacement. What is the probability that the second card is red, given that the first card is red? 40. A survey of 1000 people was conducted and the respondents were asked whether or not they smoke. The results are summarized in the following table: Male Female Smoker 127 66 Non-smoker 403 404 Find the probability that: A. The person surveyed is a smoker. B. The person surveyed is male and a non- smoker. C. The person surveyed is female and a smoker. D. The person is female, given that the person is a non-smoker. E. The person is male, given that the person is a smoker. Previous question Next question

You shuffle a standard 52-card deck, and you deal yourself two cards. Write each of the indicated answers as a fraction. What is the probability both cards are aces? What is the probability you deal yourself a pair (two cards of the same rank)? What is the probability you deal yourself two cards of the same suit? ok I figured out the first 2 but the last one is still giving me trouble $\begingroup$ As for the two cards both being the same suit... there are the two ways of phrasing it. In the one way of phrasing, you might say "oh, well, if I draw the cards in sequence, the first card can be whatever... I don't care yet what it is. Then, the next card will have to be from the same suit as the first, whatever that was" In the other phrasing you might say "pick what suit, pick what two cards from that suit, and compare that to the number of ways of having picked two cards overall." $\endgroup$ 1) The probability of drawing two aces: The first card will need to be an ace which occurs with probability $\frac$ These can be seen to equal exactly what we had before. What is the probability both cards are aces? What is the probability you deal yourself a pair (two cards of the same rank)? What is the probability you deal yourself two cards of the same suit? ok I figured out the first 2 but the last one is still giving me trouble Why? The second and third are evaluated in the same manner; just swapping suite and rank. There are 13 ranks of 4 suits, and 4 suits of 13 ranks. Count way...

A standard deck of cards has 52 members consisting of 4 suits each with 13 members. Five cards are dealt from the randomly mixed deck. What is the probability that all cards are the same suit? EDIT: How I went about it before posting this question was doing (1/4) as the first card probability because my thought process was that we'll draw 1 suit out of the 4 for the first probability. Then I proceeded to account of the 2nd dealt card with the probability of (12/51) since 1 card has been dealt already out of the 13 cards for that suit, also subtracting 1 from the total amount of cards able to be dealt. So for the 3rd card: (11/50) 4th card: (10/49) 5th card: (9/48) Giving us the total overall probability for drawing 5 cards of the same suit: $$ (1/4) * (12/51) * (11/50) * (10/49) * (9/48) = 33/66640 $$ EDIT2: My practice quiz given by TA's is still saying I have the incorrect answer. Given how the answer should be: $ 33/16660 $ (explained in numerous ways in the thread), I contacted the TA's to see if maybe the have setup the question incorrectly. Will update when I get an answer back. EDIT3: Got an answer back from my TA's who tested the test. They did have the answer wrong on their end. Everyone who helped me was correct! $\begingroup$ How many ways are there of drawing five cards from the same suit? Note that there are four suits, so the number of ways of drawing five cards from the same suit is four times, say, the number of ways of drawing five clubs. And how many ways...