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Questions about how to figure out the probability of picking from a deck of cards common in basic stats courses. For example, the probability of choosing one card, and getting a certain number card (e.g. a 7) or one from a certain suit (e.g. a club). Watch the video for examples: Probability of getting cards from a deck Watch this video on YouTube. Can’t see the video? Click here. You might wonder why you’re learning about cards (what’s the point?). The answer is that finding probabilities (like the probability of contracting an illness) can be a tricky concept to grasp at first. So your instructor will try and simplify problems using cards, dice or Bingo numbers. Once you’ve grasped the basics, you’ll start to use “real life” data for probability (usually a bit later on in the class, for example in normal distributions). Probability of picking from a deck of cards: Steps
That’s it! Tip: It isn’t as easy as just adding the number of sevens (4) and the number of clubs (13). If you did this for this example, you’d get 17 cards, not the correct answer of 16. The reason for this is that one of the cards in our example is both a club AND a number 7. Probability of picking from a deck of cards: Using ExcelWatch the video for an overview and examples of using the hypergeometric distribution in Excel for card probabilities: Excel Hypergeometric distribution to calculate card probabilities Watch this video on YouTube. Can’t see the video? Click here. It gets a LOT more complex if you’re playing a card game, you have a certain number of cards in your hand, and you want to know your odds of getting a certain card if you are drawing a certain number of cards. You have to use something called a hypergeometric distribution to figure out the odds. The formula is: H (n) = C (X, n) * C (Y – X, Z – n) / C (Y, Z) Where: X is the number of a certain card in the deck Y is the total number of cards in the deck Z is the number of cards drawn N is the number you are checking for As you can see, the formula uses combinations and factorials —it can get a bit messy to do this by hand, so consider using technology like Excel. The command in Excel is: “=HYPGEOMDIST(N,Z,X,Y)”. For example, if you have a standard 52 card deck and draw 4 cards, what will be your chances of not drawing an ace? X is 4 Y is 52 Z is 4 N is 0 (as you want zero aces!) the formula would be:
ReferencesBeyer, W. H. CRC Standard Mathematical Tables, 31st ed. Boca Raton, FL: CRC Press, pp. 536 and 571, 2002.
Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free! Comments? Need to post a correction? Please Contact Us. Sadie C. find the probability of selecting two 10s when two cards are drawn without replacement from a standard deck of cards 3 Answers By Expert Tutors
Patrick L. answered • 08/14/20 Algebra 2/Trigonometry taken in 10th Grade
A = picking a first 10 from a standard deck of cards B = picking a second 10 from a standard deck of cards There are 52 cards in a standard deck. There are 4 cards that have a "10". P(A ∩ B) = P(A)*P(B) = (4/52)*(3/51) = 0.004525 The probability of selecting two 10s from a standard deck of cards is about 0.004525.
Jon S. answered • 08/14/20 Patient and Knowledgeable Math and English Tutor
There are 4 10's in a 52 card deck. The probability of drawing the first 10 is 4/52 = 1/13, leaving 51 cards and 3 10's. The probability of drawing the second 10 is 3/51 = 1/17. The probability of both events happening is 1/13 * 1/17 = 1/221.
Cristian M. answered • 08/14/20 MS Statistics Candidate Offers Patient, Explorative and Clear Tutoring
Question: Find the probability of selecting two 10s when two cards are drawn without replacement from a standard deck of cards Answer: Remember that we have 52 cards, four of which have "10" on them. Also, remember than in sampling without replacement, our sample size will decrease. Probability of getting a 10-card: 4/52. Great. But what happens when we remove a card, and it happens to be a 10? Now there will be three 10-cards left in a stack of 51 cards. (4/52)(3/51) = 1/221. |