- **Advanced Statistics**
(*http://mymathforum.com/advanced-statistics/*)

- - **Probability of drawing a set of cards given several different drawing parameters**
(*http://mymathforum.com/advanced-statistics/46654-probability-drawing-set-cards-given-several-different-drawing-parameters.html*)

Probability of drawing a set of cards given several different drawing parametersHello - thank you in advance for any advice you can give me about this problem. First, let me tell you what I have figured out. My first problem was to figure out the probability of drawing at least one Ace from a deck of cards, given 7 draws from the deck (without returning the cards to the deck). 1. I believe the equation is: P(at least one Ace)=1-C(48,7)/C(52,7) (is this correct?) 2. I also think that if you were returning the drawn cards to the deck each time, the equation would be: P(at least one Ace)=1-(48/52)^7 (again, is this correct?) Now it starts getting more difficult. How do I figure out the probability of drawing at least 2 Aces from the deck (or at least 3 Aces), in the same 7 draws? 3. I think the equation would be: P(at least two Aces)=1-C(48,7)/C(52,7)*C(48,6)/C(51,6) (is that correct? the second one has fewer possible draws so I'm using 6 instead of 7, and there are fewer possible cards in the deck, so I'm using 51 instead of 52, but there are the same number of non-Aces if I successfully drew one Ace, so I'm still using 48 ) Now for the final question. I'm still drawing 7 times, but instead of drawing a single card each time, I'm drawing 2 cards each time, and each time I get to keep one card and must discard 1 card. Now I assume calculating the probability of getting at least one Ace is still the same, using equation 1 above (but with 14 cards rather than 7). But how does it change the probability of getting at least 2 Aces, or at least 3 Aces? 4. This would be equation 4 if I had any idea how to go about constructing it. If equation 3 above is correct, then here's my first stab: P(at least two Aces)=1-C(48,14)/C(52/14)*C(47,12)/C(51,12) (I just know this is wrong. I think the 14 is right, since we're drawing 14 total cards. But I'm not sure where to include the probability that the second card in my first draw is an Ace I had to discard, or when I calculate at least 3 Aces where I go from there) Any help would be greatly appreciated! |

Ho Kinead, at least for the first question I think you are right. When I have doubts about the solution of a particular problem I use to simulate the problem. As you will see, the aproximate solution from the simulation is very close to your answer. I think you can do something similar for the other questions. Good look. The following code is written in R (open source), I hope it is useful for you. Code: `> library(prob)` |

Hello, Kinead! Quote:
Quote:
$P(\text{at least 2 Aces}) \:=\:1 - P(\text{no Aces}) - P(\text{one Ace})$ |

All times are GMT -8. The time now is 02:12 AM. |

Copyright © 2019 My Math Forum. All rights reserved.