Probability of Picking an Ace from 47 Cards

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Homework Help Overview

The discussion revolves around calculating the probability of drawing an Ace from a modified deck of cards after removing certain cards. The subject area is probability, specifically focusing on conditional probabilities based on previous draws.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of drawing cards sequentially and how the outcome of the first draw affects the probabilities in the second draw. Questions arise regarding whether the final answer is a single probability or if it varies based on the outcome of the first draw.

Discussion Status

Participants are actively engaging with the problem, discussing the need to consider different cases based on the first draw. There is a recognition that both conditional probabilities must be calculated and combined to arrive at a final answer.

Contextual Notes

There is a moderator's note indicating that the original poster should provide an attempt at solving the problem before receiving further assistance, which suggests a focus on the learning process.

lazyp
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Pick one card from a deck, then throw it away. Now pick another card from the remaining 51 cards. What is the probability that card is an Ace?

Pick 5 cards from a deck and throw them away without looking at them. Now pick another card from the remaining 47 cards. What is the probability that card is an Ace?
 
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lazyp said:
Pick one card from a deck, then throw it away. Now pick another card from the remaining 51 cards. What is the probability that card is an Ace?

Pick 5 cards from a deck and throw them away without looking at them. Now pick another card from the remaining 47 cards. What is the probability that card is an Ace?

Is this a homework question?

When doing probability problems the best thing to do is to get a concise definition of the state space.

For example in the first experiment you pick one card and since there is 52 cards the probability of picking an ace in that experiment is 4/52 = 1/13

With the second experiment you will have probabilities that are dependent on the results of your experiment. If you pick up an ace in the first then there will be three aces in the second experiment. If there is no ace in the first trial then there there is one less "non-ace" card.

With these hints, can you formulate the probabilities of P(Get an ace | picked up an ace in first experiment) and P(Get an ace | No ace in first experiment)?
 
Yea i understand what you mean, but is the final answer to the 1st question one single probability or are there two answers(depending on whether or not an ace is picked in the 1st draw)?
 
lazyp said:
Yea i understand what you mean, but is the final answer to the 1st question one single probability or are there two answers(depending on whether or not an ace is picked in the 1st draw)?

You will end up with one final probability (answer) but you will have to consider all cases and that includes handling both of the conditional cases specified above.

Since the two cases are disjoint (mutually exclusive), you have to find both of the probabilities and add them together to get your final answer.
 
Moderator's note: thread moved to Homework & Coursework Questions from Set Theory, Logic, Probability, Statistics.

Please let the OP show an attempt at solving the problem before offering further hints and help.

Regards,

Redbelly98
 

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