Stats question help

lazyp
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?

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)?

lazyp
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)?

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.

Staff Emeritus